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Chapter 1
Vector Algebra
1.1 Rectangular Coordinates in 1-Space, 2-Space and3-Space
The Euclidean Geometry is the study of geometric elements (points, lines,planes, etc), relations between elements and configurations. The simplestgeometric elements and basic relations between them were introduced for thefirst time in an axiomatic way by Euclid (IVth-IIIth centuries B.C.), in hisfamous book ”Elements”.
A complete axiomatic system in geometry was given by D.Hilbert (1862-1943) in ”The Fundamentals of Geometry” in 1899. Two other axiomaticsystems in Euclidean geometry were introduced by G.D.Birkhoff (1884-1944)in 1932, and by H.Weyl (1885-1955) in 1917.
The basic axioms and relations are well-known from the Mathematics insecondary school or high school. They define the Euclidean plane E2 and theEuclidean space E3.
The Analytic Geometry is the study of geometric configurations by usingthe coordinates method, which works not only in E2 or E3, but also in a n-dimensional Euclidean space.
1.1.1 The Space of Coordinates Rn
Let Rn be the set of all ordered n-tuples of real numbers, i.e.
Rn = R× . . .× R.
Therefore, an element x of Rn has the form x = (x1, . . . , xn); the real numbersx1, . . ., xn are called components of x. Recall that two n-tuples (x1, . . . , xn)
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and (y1, . . . , yn) are equal if and only if their components are respectivelyequal:
(x1, . . . , xn) = (y1, . . . , yn) ⇐⇒ x1 = y1, . . . , xn = yn.
1.1.2 Cartesian Coordinates on E1
Let E1 be the 1-dimensional Euclidean space, i.e. a line d together withone of the above mentioned system of axioms. Let O and A be two differentpoints, fixed on d, such that OA = 1 (see Figure 1.1).
Figure 1.1:
One obtains an orientation on the line d (from O to A) and one can intro-duce the function
f1 : E1 → R, f1(P ) = xP ,
where |xP | = OP and{xP ≥ 0 if P ∈ [OAxP < 0 if P /∈ [OA
. Hence, f is bijective and one
can associate to any point P ∈ E1 a unique real number xP . One says thatOx is a Cartesian system of coordinates on E1, having the origin O and theaxis Ox, while xP is said to be the coordinate of P . We shall use the notationP (xP ).
1.1.3 Rectangular Coordinates on E2
Let E2 be the 2-dimensional Euclidean space, i.e. a plane π together witha system of axioms mentioned at the beginning of this chapter. Let O ∈ E2
be a fixed point d and d′ be two orthogonal lines, passing through O. Onecan choose, on each of the lines d and d′, a Cartesian system of coordinates,having the same origin O. Suppose they are denoted by Ox respectively Oy.
If P is an arbitrary point of E2, let xP and yP be the coordinates of theorthogonal projections Px and Py on Ox respectively Oy. One can associate toany point P ∈ E2 a unique pair (xP , yP ) ∈ R2, so that one obtains a bijection
f2 : E2 → R2, f2(P ) = (xP , yP ).
The origin O together with the axes Ox and Oy form a Cartesian system ofcoordinates Oxy on E2, while the rectangular coordinates of P in this systemare (xP , yP ) (see Figure 1.2).
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Figure 1.2:
The axes d and d′ divide the plane E2 into four domains which are namedquadrants and are denoted by I, II, III, IV. One can characterize to whichquadrant of the plane a point P belongs, by looking at the signs of coordinates(xP , yP ) of the point P (which is not situated on the coordinates lines Ox,Oy).
Quadrant I II III IVxP + − − +yp + + − −
A simple replacement in Pythagora’s Theorem shows that the length ofthe segment [P1P2], where P1(x1, y1), P2(x2, y2), is given by the formula
P1P2 =√
(x2 − x1)2 + (y2 − y1)2. (1.1)
If the point P divides the segment [P1P2] into the ratio k, i.e.PP1
PP2= k,
then the coordinates of P are
P
(x1 + kx2
1 + k,y1 + ky2
1 + k
). (1.2)
1.1.4 Rectangular Coordinates in E3
Let E3 be the 3-dimensional Euclidean space, O be a fixed point and d, d′,d′′ be three pairwise orthogonal lines, passing through O. Choose, on the linesd, d′ and d′′, the Cartesian systems of coordinates Ox, Oy respectively Oz,having the same origin O (see Figure 1.3).
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For an arbitrary point in P ∈ E3, denote by xP , yP and zP the coordinatesof its orthogonal projections Px, Py and Pz on d, d′ respectively d′′. One candefine the bijection
f3 : E3 → R3, f3(P ) = (xP , yP , zP ).
Figure 1.3:
One introduced on E3 the right rectangular coordinates system (or theCartesian system) Oxyz, having the following elements:
• the origin O;
• the coordinate lines (or axes) Ox, Oy, Oz;
• the coordinate planes Oxy, Oyz, Ozx;
Examples.1) The origin O has the coordinates (0, 0, 0).2) The points situated on Ox, Oy and Oz are of coordinates (x, 0, 0),
(0, y, 0), respectively (0, 0, z).3) The points situated on Oxy, Oyz and Ozx have the coordinates (x, y, 0),
(0, y, z) respectively (x, 0, z).The three coordinate planes divide the space E3 into eight domains, de-
noted by I, II, ..., VIII. These domains are defined by the signs of the coordi-nates of the points they contain, as in the following table:
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Domain I II III IV V VI VII VIIIxP + − − + + − − +yP + + − − + + − −zP + + + + − − − −
Theorem 1.1.4.1. (the distance formula) The distance between the pointsP1(x1, y1, z1) and P2(x2, y2, z2) is given by
P1P2 =√
(x2 − x1)2 + (y2 − y1)2 + (z2 − z1)2. (1.3)
Proof: Construct a right parallelepiped having P1 and P2 as opposite ver-tices and the faces parallel to the coordinate planes, as in Figure 1.4. One
Figure 1.4:
hasP1A = |y2 − y1|,
AB = |x2 − x1|,
BP2 = |z2 − z1|.
SinceP1B
2 = P1A2 +AB2 = (y2 − y1)2 + (x2 − x1)2,
then
P1P22 = P1B
2 +BP 22 = (x2 − x1)2 + (y2 − y1)2 + (z2 − z1)2,
and the formula 1.3 is obtained. �
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Theorem 1.1.4.2. If the point P divides the segment [P1P2], P1(x1, y1, z1),
P2(x2, y2, z2), into the ratio k (i.e.PP1
PP2= k), then the coordinates of P are
(x1 + kx2
1 + k,y1 + ky2
1 + k,z1 + kz2
1 + k
). (1.4)
Proof: Take P (xP , yP , zP ) and construct PP ′||P1A, with P ′ ∈ AP2 (seeFigure 1.4). The similarity of the triangles ∆PP ′P2 and ∆P1AP2 yields
PP ′
P1A=
PP2
P1P2=
1k + 1
,
which is equivalent to|y2 − yP ||y2 − y1|
=k
k + 1.
Supposing that y2 > y1 and yP > y1 (otherwise, the calculations are analo-
gous), one obtains yP =y1 + ky2
1 + k.
Similarly, xP =x1 + kx2
1 + kand zP =
z1 + kz2
1 + k. �
Remark. If M is the midpoint of the segment [P1P2], determined byP1(x1, y1, z1) and P2(x2, y2, z2), then k = 1 and the coordinates of M are
M
(x1 + x2
2,y1 + y2
2,z1 + z2
2
). (1.5)
1.1.5 Exercises
1. Give the coordinates of the vertices of the rectangular parallelepipedwhose sides are the coordinate planes and the planes x = 1, y = 3,z = 6.
2. Describe the locus of points P (x, y, z) in each of the following situations:
a) xyz = 0;
b) x2 + y2 + z2 = 0;
c) (x+ 1)2 + (y − 2)2 + (z + 3)2 = 0;
d) (x− 2)(z − 8) = 0;
e) z2 − 25 = 0;
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3. Show that the given points are collinear:
a) P1(1, 2, 9), P2(−2,−2,−3), P3(7, 10, 6);
b) Q1(2, 3, 2), Q2(1, 4, 4), Q3(5, 0,−4).
4. Find x if:
a) P1(x, 2, 3), P2(2, 1, 1) and P1P2 =√
21;
b) Q1(x, x, 1), Q2(0, 3, 5) and Q1Q2 = 5.
5. The coordinates of the midpoint of the segment [P1P2], determined byP1(x1, y1, z1) and P2(2, 3, 6), are (−1,−4, 8). Find the coordinates of P1.
6. Let P3 be the midpoint of the segment joining the points P1(−3, 4, 1)and P2(−5, 8, 3). Find the coordinates of the midpoint of the segment:
a) joining P1 and P3;
b) joining P3 and P2.
1.2 Other Coordinate Systems in E2 and E3
Up to now, we used a rectangular coordinate system to specify a point P inthe plane E2 or in the space E3. There are many situations when the calcula-tions are much simplified by introducing some different coordinate systems.
1.2.1 The Polar Coordinate System (PS)
As an alternative to a rectangular coordinate system (RS) one considers inthe plane E2 a fixed point O, called pole and a half-line directed to the rightof O, called polar axis (see Figure 1.5).
Figure 1.5:
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By specifying a directed distance ρ from O to a point P and an angleθ (measured in radians), whose ”initial” side is the polar axis and whose”terminal” side is the ray OP , the polar coordinates of the point P are (ρ, θ).
One obtains a bijection
E2 \ {O} → R+ × [0, 2π), P → (ρ, θ)
which associates to any point P in E2 \ {O} the pair (ρ, θ) (suppose thatO(0, 0)). The positive real number ρ is called the polar ray of P and θ iscalled the polar angle of P .
Consider RS to be the rectangular coordinate system in E2, whose originO is the pole and whose positive half-axis Ox is the polar axis (see Figure1.6). The following transformation formulas give the connection between thecoordinates of an arbitrary point in the two systems of coordinates.
PS→ RS Let P (ρ, θ) be a point in the system PS. It is immediate that{xP = ρ cos θyP = ρ sin θ
. (1.6)
Figure 1.6:
RS → PS Let P (x, y) be a point in the system RS. It is clear that thepolar ray of P is given by the formula
ρ =√x2 + y2. (1.7)
In order to obtain the polar angle of P , it must be considered the quadrantwhere P is situated. One obtains the following formulas:
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Case 1. If x 6= 0, then using tan θ =y
x, one has
θ = arctany
x+ kπ, where k =
0 if P ∈ I ∪ (Ox1 if P ∈ II ∪ III ∪ (Ox′
2 if P ∈ IV;
Case 2. If x = 0 and y 6= 0, then
θ =
π
2when P ∈ (Oy
3π2
when P ∈ (Oy′;
Case 3. If x = 0 and y = 0, then θ = 0.
1.2.2 The Cylindrical Coordinate System (CS)
In order to have a valid coordinate system in the 3-dimensional case, eachpoint of the space must be associated to a unique triple of real numbers (thecoordinates of the point) and each triple of real numbers must determine aunique point, as in the case of the rectangular system of coordinates.
Let P (x, y, z) be a point in a rectangular system of coordinates Oxyz andP ′ be the orthogonal projection of P on the plane xOy. One can associate tothe point P the triple (r, θ, z), where (r, θ) are the polar coordinates of P ′ (seeFigure 1.7).
Figure 1.7:
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The triple (r, θ, z) gives the cylindrical coordinates of the point P . Thereis the bijection
h1 : E3 \ {O} → R+ × [0, 2π)× R, P → (r, θ, z)
and one obtains a new coordinate system, named the cylindrical coordinatesystem (CS) in E3.
In the following table, the conversion formulas relative to the cylindricalcoordinate system (CS) and the rectangular coordinate system (RS) are pre-sented.
Conversion FormulasCS→RS x = r cos θ, y = r sin θ, z = z
(r, θ, z) → (x, y, z)RS→CS r =
√x2 + y2, z = z and θ is given as follows:
(x, y, z) → (r, θ, z) Case 1. If x 6= 0, then
θ = arctany
x+ kπ,
where k =
0, if P ∈ I ∪ (Ox1, if P ∈ II ∪ III ∪ (Ox′
2, if P ∈ IVCase 2. If x = 0 and y 6= 0, then
θ =
π
2when P ∈ (Oy
3π2
when P ∈ (Oy′
Case 3. If x = 0 and y = 0, then θ = 0.
Examples1) In the cylindrical coordinate system, the equation r = r0 represents a
right circular cylinder of radius r0, centered on the z-axis.2) The equation θ = θ0 describes a half-plane attached along the z-axis
and making an angle θ0 with the positive x-axis.3) The equation z = z0 defines a plane which is parallel to the coordinate
plane xOy.
1.2.3 The Spherical Coordinate System (SS)
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Another way to associate to each point P in E3 a triple of real numbersis illustrated in Figure 1.8. If P (x, y, z) is a point in a rectangular system ofcoordinates Oxyz and P ′ its orthogonal projection on Oxy, let ρ be the lengthof the segment [OP ], θ be the oriented angle determined by [Ox and [OP ′ andϕ be the oriented angle between [Oz and [OP . The triple (ρ, θ, ϕ) gives the
Figure 1.8:
spherical coordinates of the point P . This way, one obtains the bijection
h2 : E3 \ {O} → R+ × [0, 2π)× [0, π], P → (ρ, θ, ϕ),
which defines a new coordinate system in E3, called the spherical coordinatesystem (SS).
The conversion formulas involving the spherical coordinate system (SS)and the rectangular coordinate system (RS) are presented in the followingtable.
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Conversion FormulasSS→RS x = ρ cos θ sinϕ, y = ρ sin θ sinϕ, z = ρ cosϕ
(ρ, θ, ϕ) → (x, y, z)
RS→SS ρ =√x2 + y2 + z2, ϕ = arccos
z√x2 + y2 + z2
(x, y, z) → (ρ, θ, ϕ) θ is given as follows:Case 1. If x 6= 0, then
θ = arctany
x+ kπ,
where k =
0, P ′ ∈ I ∪ (Ox1, P ′ ∈ II ∪ III ∪ (Ox′
2, P ′ ∈ IVCase 2. If x = 0 and y 6= 0, then
θ =
π
2, P ′ ∈ (Oy
3π2, P ′ ∈ (Oy′
Case 3. If x = 0 and y = 0, then θ = 0
Examples1) In the spherical coordinate system, the equation ρ = ρ0 represents the
set of all points in E3 whose distance ρ to the origin is ρ0. This is a sphere ofradius ρ0, centered at the origin.
2) As in the cylindrical coordinates, the equation θ = θ0 defines a half-planeattached along the z-axis, making an angle θ0 with the positive x-axis.
3) The equation ϕ = ϕ0 describes the points P for which the angle deter-
mined by [OP and [Oz is ϕ0. If ϕ0 6=π
2and ϕ0 6= π, this is a right circular
cone, having the vertex at the origin and centered on the z-axis. The equation
ϕ =π
2defines the coordinate plane xOy. The equation ϕ = π describes the
negative axis (Oz′.
1.2.4 Exercises
1. Graph the point P whose polar coordinates are given by:a)(2, π); b)(3, π/3); c)(4, 3π/2); d)(5, π/6).
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2. Find the rectangular coordinates of the points whose polar coordinatesare given by:a)(1, 2π/3); b)(1/2, 7π/4); c)(7, π/3); d)(
√3, 11π/6).
3. Find the polar coordinates of the points whose rectangular coordinatesare given by:a)(−3,−3); b)(0,−5); c)(
√3,−1); d)(
√2,√
6).
4. Find the polar equation corresponding to the given Cartesian equation:
a) y = 5
b) x+ 1 = 0
c) y = 7x
d) 3x+ 8y + 6 = 0
e) y2 = −4x+ 4
f) x2 − 12y − 36 = 0
g) x2 + y2 = 36
h) x2 − y2 = 25.
5. Determine, in cylindrical coordinates, the equation of the surface whoseequation in rectangular coordinates is
z = x2 + y2 − 2x+ y.
6. Find the equation, in rectangular coordinates, of the surface whose equa-tion in cylindrical coordinates is r = 4 cos θ.
7. Find the equation, in spherical coordinates, of the surface whose equationin rectangular coordinates is
z = x2 + y2.
8. Express, in rectangular and spherical coordinates, the following equa-tions, given in cylindrical coordinates:
a) r2 + z2 = 1;
b) θ =π
4;
c) r2 cos 2θ = z.
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9. The equations below are given in spherical coordinates. Express themin rectangular coordinates:
a) ρ sinϕ = 2 cos θ;b) ρ− 2 sinϕ cos θ = 0.
10. Express, in cylindrical and spherical coordinates, the following equationsgiven in rectangular coordinates:
a) z = 3x2 + 3y2;b) x2 + y2 + z2 = 2z.
1.3 Vectors
Let E denote the Euclidean plane E2 or the Euclidean 3-space E3. A pair(A,B) ∈ E × E is called an ordered pair of points or a vector at the point A.Such a pair is, shortly, denoted by
−−→AB. The point A is the original point, while
B is the terminal point and the line AB (if A 6= B) gives the direction of−−→AB.
A vector−−→AB at A has the orientation from A to B, i.e. from its original to
its terminal point. The length of the segment [AB] represents the length ofthe vector
−−→AB and is denoted by ||
−−→AB|| or by |
−−→AB|. Usually, the vector
−−→AB
at A is represented as in Figure 1.9. For A = B, one obtains the zero vector
Figure 1.9:
at the point A.Define the following relation on E × E : (A,B) ∼ (C,D) if and only if the
segments [AD] and [BC] have the same midpoint. When the points A, B, Cand D are not collinear, this means that (A,B) ∼ (C,D) if and only if ABCDis a parallelogram.
It is not difficult to check that ” ∼ ” is an equivalence relation. Let usdenote by V3 the set (E3 × E3)/∼ of equivalence classes and by V2 the set(E2 × E2)/∼.
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If−−→AB ∈ E×E , its equivalence class is denoted by AB and is called a vector
in E (E2 or E3). In this case,−−→AB is a representer of AB.
Suppose that A 6= B. The line AB defines the direction of the vector AB.The length of AB is given by
||AB|| = ||−−→AB|| = AB,
the length of the segment [AB]. The orientation of AB, from A to B, is givenby the orientation of
−−→AB.
If A = B, denote by 0 the zero vector represented by−→AA.
Generally, we shall denote the vectors in V2 or V3 by small letters: a,b,. . .u, v, w.
Proposition 1.3.1. Given a vector a in V2 (or V3) and a fixed point A, thereexists a unique representer of a, having the original point at A.
Proof: Choose an arbitrary representer−−→CD ∈ a and consider the line
d||CD, passing through A. Taking into account the orientation, there exists a
Figure 1.10:
unique point B such that−−→AB ∼
−−→CD (see Figure 1.10). �
1.3.1 Vector Operations. Components
Let a and b be two vectors in V3 (or V2). The sum of a and b is the vectordenoted by a+ b, so that, if
−−→AB ∈ a and
−−→BC ∈ b, then
−→AC is the representer
of a+ b (see Figure 1.11).If v is a vector in V3 (or V2), then the opposite vector of v is denoted by
−v, so that, if−−→AB is a representer of v, then
−−→BA is a representer of −v (see
Figure 1.12).The sum a + (−b) will be, shortly, denoted by a − b and it will be called
the difference of the vectors a and b.Let a be a vector in V3 (or V2) and k be a real number. The product k · a
is the vector defined as follows:
• 0 if a = 0 or k = 0;
25
Figure 1.11:
Figure 1.12:
• if k > 0, then k · a has the same direction and orientation as a and||k · a|| = k · ||a||;
• if k < 0, then k · a has the same direction as a, opposite orientation toa and ||k · a|| = −k · ||a||.
Let a be a vector in V2 and xOy be a rectangular coordinates system inE2. There exists a unique point A ∈ E2, such that
−→OA ∈ a (see Figure 1.13).
The coordinates of the point A are called the components of the vector a; wedenote it by a(a1, a2).
Given a a vector in V3 and a rectangular coordinate system Oxyz inE3, there exists a unique point A(a1, a2, a3), such that
−→OA ∈ a. The triple
(a1, a2, a3) gives the components of a and we denote it by a(a1, a2, a3).Since 0(0, 0) in V2 and 0(0, 0, 0) in V3, then two vectors are equal if and
only if they have the same components.
Theorem 1.3.1.1. Let a(a1, a2) and b(b1, b2) be two vectors in V2 and k ∈ R.Then:
26
Figure 1.13:
1) the components of a+ b are (a1 + b1, a2 + b2);
2) the components of k · a are (ka1, ka2).
Proof: 1) Let−→OA ∈ a and
−−→OB ∈ b be representers for a respectively b,
having the same original point at the origin of the rectangular coordinatessystem xOy (see Figure 1.14). If
−−→OC ∈ a+ b, then OBCA is a parallelogram
Figure 1.14:
and it follows that c1 = a1 + b1. In an analogous way, one can show thatc2 = a2 + b2. �
With the same argument, one obtains:
27
Theorem 1.3.1.2. Let a(a1, a2, a3) and b(b1, b2, b3) be two vectors in V3 andk ∈ R. Then:
1) the components of a+ b are (a1 + b1, a2 + b2, a3 + b3);
2) the components of k · a are (ka1, ka2, ka3).
Theorem 1.3.1.3. 1) If P1(x1, y1) and P2(x2, y2) are two points in E2, then
P1P2(x2 − x1, y2 − y1).
2) If Q1(x1, y1, z1) and Q2(x2, y2, z2) are two points in E3, then
Q1Q2(x2 − x1, y2 − y2, z2 − z1).
Proof: 2) Let O be the origin of the rectangular coordinates system andremark that
Q1Q2 = Q1O +OQ2 = OQ2 −OQ1.
Using Theorem 1.3.1.2, it follows that the vector OQ2 −OQ1 has the compo-nents (x2 − x1, y2 − y1, z2 − z1), therefore
Q1Q2(x2 − x1, y2 − y1, z2 − z1). �
Theorem 1.3.1.4. (properties of summation). Let a, b and c be vectors inV3 (or V2) and α, β ∈ R. Then:
1) a+ b = b+ a (commutativity);
2) (a+ b) + c = a+ (b+ c) (associativity);
3) a+ 0 = 0 + a = a (0 is the neutral element for summation);
4) a+ (−a) = (−a) + a = 0 (−a is the symmetrical of a);
5) α(βa) = (αβ)a;
6) α · (a+ b) = α · a+β · b (distributiveness of multiplication by real scalarswith respect to the summation of vectors);
7) (α+β) ·a = α ·a+β ·a (distributiveness of multiplication by real scalarswith respect to the summation of scalars);
8) 1 · a = a.
28
Proof: All the equalities can be proved simply by expressing the compo-nents of the vectors on the left and right sides. For instance,
2) If a(a1, a2, a3), b(b1, b2, b3) and c(c1, c2, c3), then the components of thevectors (a+b)+c and a+(b+c) are ((a1 +b1)+c1, (a2 +b2)+c2, (a3 +b3)+c3)respectively (a1 + (b1 + c1), a2 + (b2 + c2), a3 + (b3 + c3)) and the conclusionfollows. �
Proposition 1.3.1.5. 1) Let a(a1, a2) be a vector in V2. The length of a isgiven by
||a|| =√a2
1 + a22.
2) Let a(a1, a2, a3) be a vector in V3. The length of a is given by
||a|| =√a2
1 + a22 + a3
3.
Proof: 2) If−→OA is the representer of a having the original point at the
origin of the rectangular coordinate system, then O(0, 0, 0), A(a1, a2, a3) and
||a|| = ||−→OA|| =
√(a1 − 0)2 + (a2 − 0)2 + (a3 − 0)2 =
√a2
1 + a22 + a2
3. �
• The vectors i(1, 0) and j(0, 1) in V2 are called the unit vectors (or ver-sors) of the coordinate axes Ox and Oy.
• The vectors i(1, 0, 0), j(0, 1, 0) and k(0, 0, 1) are called the unit vectors(or versors) of the coordinate axes Ox, Oy and Oz (see Figure 1.15).
It is clear that||i|| = ||j|| = ||k|| = 1.
We have defined the operations
+ : V2 × V2 → V2, (a, b) 7→ a+ b
· : R× V2 → V2, (k, a) 7→ k · a
respectively+ : V3 × V3 → V3, (a, b) 7→ a+ b
· : R× V3 → V3, (k, a) 7→ k · a.
The following result points out the important algebraic structure of V2 andV3.
29
Figure 1.15:
Theorem 1.3.1.6. 1) (V2,+, ·) is a vector space over R, which is isomorphicto (R2,+, ·). The set {i, j} is a base of V2, therefore dimRV2 = 2.
2) (V3,+, ·) is a vector space over R, which is isomorphic to (R3,+, ·). Theset {i, j, k} is a base of V3, therefore dimRV3 = 3.
Proof: 2) Theorem 1.3.1.4 contains the axioms of a vector space. Themap ψ : V3 → R3, given by ψ(a) = (a1, a2, a3), where a has the components(a1, a2, a3), is bijective and it satisfies
ψ(αa+ βb) = αψ(a) + βψ(b), ∀ a, b ∈ V3 and α, β ∈ R.
Indeed, if a(a1, a2, a3) and b(b1, b2, b3), one has
αa+ βb(αa1 + βb1, αa2 + βb2, αa3 + βb3),
so thatψ(αa+ βb) = (αa1 + βb1, αa2 + βb2, αa3 + βb3) =
= α(a1, a2, a3) + β(b1, b2, b3) = αψ(a) + βψ(b).
Therefore, the map ψ is an isomorphism between V3 and R3.It is well known that {e1, e2, e3} is a base in R3, where e1(1, 0, 0), e2(0, 1, 0)
and e3(0, 0, 1). Since ψ(i) = e1, ψ(j) = e2, ψ(k) = e3 and ψ is an isomorphism,then {i, j, k} is a base in V3. �
The base {i, j, k} is called the canonical base of V3.Remark : The vector space V2 is isomorphic to the 2-dimensional subspace
of V3 given byS = {a(a1, a2, a3) : a3 = 0}
and an isomorphism is φ : V2 → S, φ(a) = a′, where a(a1, a2) and a′(a1, a2, 0).
30
• Let a and b be two nonzero vectors in V3 (or V2). They are linearlydependent if there exist the scalars α, β ∈ R∗ such that αa+ βb = 0.
• Let set a, b and c be three nonzero vectors in V3. They are linearlydependent if there exist the scalars α, β, γ ∈ R, not all equal to zero,such that αa+ βb+ γc = 0.
• The vectors a and b in V3 (or V2), a, b 6= 0, are collinear if they haverepresenters situated on the same line.
• The vectors a, b and c in V3, a, b, c 6= 0 are coplanar if they have repre-senters situated in the same plane.
Theorem 1.3.1.7. 1) The vectors a and b are linearly dependent if and onlyif they are collinear.
2) The vectors a, b and c are linearly dependent in V3 if and only if theyare coplanar.
Proof: 1) If the vectors a and b are collinear, then there exists a scalar α ∈ R∗such that a = α · b, i.e.
1 · a+ (−α) · b = 0,
so that a and b are linearly dependent.
Conversely, if αa+ βb = 0 for some scalars α and β, then a =
(−β
α
)b, so
that a and b are collinear.2) Suppose that the vectors a, b and c are linearly dependent. Then, there
exist α, β, γ ∈ R not all zero, such that αa+βb+ γc = 0. Suppose that γ 6= 0.One obtains
c =
(−α
γ
)a+
(−β
γ
)b.
If−→OA and
−−→OB are representers of a respectively b, then the representer
−−→OC
of c, constructed as in Figure 1.16, is coplanar with−→OA and
−−→OB.
Conversely, if a, b and c are coplanar, let us consider the representers−→OA ∈ a,
−−→OB ∈ b and
−−→OC ∈ c, situated in the same plane. Let OMCN be the
parallelogram constructed as in Figure 1.17. Then, there exist α, β ∈ R suchthat
−−→OM = α·
−→OA and
−−→ON = β ·
−−→OB. Hence OC = OM+ON = α·OA+β ·OB
and c = α · a+ β · b, so that α · a+ β · b+ (−1) · c = 0 and the vectors a, b andc are linearly dependent. �
31
Figure 1.16:
Figure 1.17:
Corollary 1.3.1.8. 1) The set {a, b} is a base in V2 if and only if the vectorsa, b are not collinear.
2) The set {a, b, c} is a base in V3 if and only if the vectors a, b, c are notcoplanar.
1.3.2 Dot product. Projections
One defines the angle determined by two nonzero vectors a and b from V2
(or V3), θ = (a, b), to be the angle determined by their directions, taking intoaccount their orientations, such that θ ∈ [0, π] (see Figure 1.18)
Figure 1.18:
Now, given the vectors a and b in V2 (or V3), their dot product is the real
32
number defined through
a · b ={|a||b| cos θ, if a 6= 0, b 6= 0
0, otherwise. (1.8)
Theorem 1.3.2.1. 1) If a(a1, a2) and b(b1, b2) are two vectors in V2, then
a · b = a1b1 + a2b2; (1.9)
2) If a(a1, a2, a3) and b(b1, b2, b3) are two vectors in V3, then
a · b = a1b1 + a2b2 + a3b3. (1.10)
Proof: Let−→OA and
−−→OB be representers of a, respectively b, with the same
original point O (see Figure 1.19). The cosine theorem in the triangle ∆OAB
Figure 1.19:
gives:AB2 = |a|2 + |b|2 − 2|a||b| cos θ,
hence
a · b = |a||b| cos θ =12(|a|2 + |b|2 −AB2) =
12(|a|2 + |b|2 − |AB|2) =
=12[(a2
1 + a22 + a2
3) + (b21 + b22 + b23)− (b1 − a1)2 − (b2 − a2)2 − (b3 − a3)2] =
= a1b1 + a2b2 + a3b3. �
Since cos θ =a · b|a||b|
, then, for two nonzero vectors a and b, one has
cos (a, b) =a1b1 + a2b2√
a21 + a2
2 ·√b21 + b22
, for a, b ∈ V2; (1.11)
cos (a, b) =a1b1 + a2b2 + a3b3√
a21 + a2
2 + a23 ·√b21 + b22 + b23
, for a, b ∈ V3. (1.12)
33
Theorem 1.3.2.2. If u and v are nonzero vectors in V2 (or V3) and θ is theangle between them, then
a) θ is acute if and only if u · v > 0;
b) θ is obtuse if and only if u · v < 0;
c) θ =π
2if and only if u · v = 0.
Proof: The sign of the cosine of the angle determined by two vectorscoincides with the sign of their dot product. The assertions are immediate. �
Given an arbitrary vector u ∈ V3 and an associated Cartesian systemof coordinates, one defines the director angles of u to be the three anglesdetermined by u with the versors of the system of coordinates, respectively:
α = (u, i), β = (u, j) and γ = (u, k).
Figure 1.20:
The values cosα, cosβ and cos γ are the director cosines of the vector u.
Theorem 1.3.2.3. The director cosines of a vector u(u1, u2, u3) ∈ V3, u 6= 0,are
cosα =u1
|u|, cosβ =
u2
|u|, cos γ =
u3
|u|. (1.13)
Proof: Since i(1, 0, 0), the formula 1.12 gives cosα =u · i|u||i|
=u1
|u|. �
34
Remark : For any nonzero vector u ∈ V3,u
|u|is a unit vector, called the
versor of u and, obviously,
u
|u|= cosα · i+ cosβ · j + cos γ · k, with (cosα)2 + (cosβ)2 + (cos γ)2 = 1.
Theorem 1.3.2.4. (algebraic properties of the dot product) Given a, b, c ∈ V3
(or V2) and λ ∈ R, one has:
1) a · b = b · a (commutativity of the dot product);
2) a · (b+ c) = a · b+ a · c (distributivity of the dot product with respect tothe summation of vectors);
3) λ(a · b) = (λa) · b = a · (λb);
4) a · a = |a|2.
Sometimes it is useful to decompose a vector into a sum of two terms,one of them having a given direction and the other being orthogonal on thisdirection.
Figure 1.21:
Let u and b be two nonzero vectors and project (orthogonally) a representerof the vector u on a line passing through the original point of this representerand parallel to the direction of b. One gets the vector w1, having the directionof b and, by making the difference u − w1, another vector w2, orthogonal onthe direction of b (see Figure 1.21); u = w1 + w2.
The vector w1 is called the orthogonal projection of u on b and it is denotedby prbu. The vector w2 is called the vector component of u orthogonal to band w2 = u−prbu.
35
Theorem 1.3.2.5. If u and b are vectors in V2 or V3 and b 6= 0, then
• the orthogonal projection of u on b is prbu =u · b|b|2
· b;
• the vector component of u orthogonal to b is u-prbu = u−u · b|b|2
· b.
Proof: Since w1 is parallel to b, there exists a real number k, such thatw1 = kb. Thus,
u = w1 + w2 = kb+ w2.
Multiplying by the vector b, one obtains:
u · b = (w1 + w2) · b = (kb+ w2) · b = k|b|2,
since w2 and b are orthogonal. Then the constant k is given by k =u · b|b|2
,
which completes the proof of the theorem. �The length of the orthogonal projection of the vector u on b can be obtained
as following:
|prbu| =∣∣∣∣u · b|b|2
· b∣∣∣∣ = ∣∣∣∣u · b|b|2
∣∣∣∣ |b|,which yields
|prbu| =|u · b||b|
. (1.14)
Remark : If θ is the angle between u and b, then
|prbu| = |u|| cos θ|.
1.3.3 Cross Product. Triple Scalar Product
If u = u1i+ u2j + u3k and v = v1i+ v2j + v3k are vectors in V3, then theircross product u× v is the vector
u× v =∣∣∣∣u2 u3
v2 v3
∣∣∣∣ i+ ∣∣∣∣u1 u3
v1 v3
∣∣∣∣ j +∣∣∣∣u1 u2
v1 v2
∣∣∣∣ k, (1.15)
or, shortly,
u× v =
∣∣∣∣∣∣i j ku1 u2 u3
v1 v2 v3
∣∣∣∣∣∣ . (1.16)
36
Theorem 1.3.3.1. If u and v are two vectors in V3, then
1) u · (u× v) = 0 (u× v is orthogonal on u);
2) v · (u× v) = 0 (u× v is orthogonal on v);
3) |u× v|2 = |u|2|v|2 − (u · v)2 (Lagrange’s identity).
Proof: Let u(u1, u2, u3) and v(v1, v2, v3). The components of u× v are
u× v(u2v3 − u3v2, u3v1 − u1v3, u1v2 − u2v1).
Hence,
u · (u× v) = u1(u2v3 − u3v2) + u2(u3v1 − u1v3) + u3(u1v2 − u2v1) = 0,
v · (u× v) = v1(u2v3 − u3v2) + v2(u3v1 − u1v3) + v3(u1v2 − u2v1) = 0,
and a simple computation will show that
|u× v| = (u2v3 − u3v2)2 + (u3v1 − u1v3)2 + (u1v2 − u2v1)2
equals to
|u|2|v|2 − (u · v)2 = (u21 + u2
2 + u23)(v
21 + v2
2 + v23)− (u1v1 + u2v2 + u3v3)2. �
Suppose that u and v are nonzero vectors in V3. An immediate consequenceof the Lagrange’s identity is that |u|2|v|2− (u ·v)2 ≥ 0, or |u ·v| ≤ |u||v|, whichleads, after replacing the components of the vectors, to the Cauchy-Schwartzinequality. The equality |u · v| = |u||v| holds if and only if the vector u× v isthe zero vector, i.e. its components are all zero, which happens if and only ifv1
u1=v2
u2=v3
u3= λ, or v = λu, λ ∈ R∗. In summary, one has:
Theorem 1.3.3.2. If u and v are nonzero vectors in V3, then u × v = 0 ifand only if u and v are parallel.
Using the Lagrange’s identity and the definition of the dot product, onecan determine the length of the cross product of two nonzero vectors from V3.If θ is the angle determined by u and v, θ ∈ [0, π], then
|u× v|2 = |u|2|v|2 − (u · v)2 = |u|2|v|2 − |u|2|v|2 cos2 θ = |u|2|v|2 sin2 θ,
hence|u× v| = |u||v| sin θ. (1.17)
The formula 1.17 has a useful geometric meaning: the length of the crossproduct of two nonzero vectors from V3 is exactly the area of the parallelogramconstructed on the two vectors (see Figure 1.22).
The vector u×v is completely determined by the following three properties:
37
Figure 1.22:
• u× v is orthogonal on both u and v (hence on the plane determined bytwo coplanar representers of u and v);
• the orientation of u× v is given by the right-hand rule;
• |u× v| = |u||v| sin θ.
Figure 1.23:
Theorem 1.3.3.3. (algebraic properties of the cross product) For any vectorsu, v and w from V3 and any scalar λ ∈ R, the following equalities hold:
a) u× v = −v × u;
b) u× (v + w) = u× v + u× w;
c) (u+ v)× w = u× w + v × w;
d) λ(u× v) = (λu)× v = u× (λv);
e) u× 0 = 0× u = 0;
38
f) u× u = 0.
It is very easy to compute the cross products of the versors of the axes:
i× j = k j × k = i k × i = j
j × i = −k k × j = −i i× k = −ji× i = 0 j × j = 0 k × k = 0
.
Given three vectors a, b and c from V3, one defines their triple scalarproduct to be the real number (a, b, c) = a · (b× c).
If a = (a1, a2, a3), b = (b1, b2, b3) and c = (c1, c2, c3), then the triple scalarproduct can be calculated as
(a, b, c) =
∣∣∣∣∣∣a1 a2 a3
b1 b2 b3c1 c2 c3
∣∣∣∣∣∣ . (1.18)
Indeed,
(a, b, c) = a1(b2c3 − b3c2) + a2(b3c1 − b1c3) + a3(b1c2 − b2c1) =
= a1
∣∣∣∣b2 b3c2 c3
∣∣∣∣+ a2
∣∣∣∣b3 b1c3 c1
∣∣∣∣+ a3
∣∣∣∣b1 b2c1 c2
∣∣∣∣ =∣∣∣∣∣∣a1 a2 a3
b1 b2 b3c1 c2 c3
∣∣∣∣∣∣ .Remark : It can be seen easily that the triple scalar product can be also
seen as (a, b, c) = (a× b) · c.
Theorem 1.3.3.4. (properties of the triple scalar product) If a, b and c arevectors in V3, then:
a) (a, b, c) = (c, a, b) = (b, c, a);
b) (a, b, c) = 0 if and only if a, b and c are linearly dependent (i.e. theyhave representers situated on the same plane).
Proof: a) It follows immediately from the properties of the determinants;b) The triple scalar product (a, b, c) = 0 if and only if the determinant∣∣∣∣∣∣
a1 a2 a3
b1 b2 b3c1 c2 c3
∣∣∣∣∣∣ = 0, therefore its rows, for instance, are linearly dependent,
meaning that the vectors a, b and c are linearly dependent, so that one canchoose a representer for each, having the same original point, which are con-tained into a plane. �
39
The triple scalar product has a geometric meaning. Suppose that the vec-tors a, b and c are linearly independent and choose a representer for each,having the same original point. These form the adjacent sides of a paral-lelepiped (see Figure 1.24). Suppose that the base of this parallelepiped is
Figure 1.24:
the parallelogram constructed on b and c. The height of the parallelepiped isthe length of the orthogonal projection of the vector a on the direction of thevector b× c,
h = |prb×ca| =∣∣∣∣a · (b× c)|b× c|
∣∣∣∣ = |(a, b, c)||b× c|
.
Then, the volume of the parallelepiped whose adjacent sides are the vectorsa, b and c is the absolute value of the triple scalar product (a, b, c):
V = h ·Area(b, c) =|(a, b, c)||b× c|
|b× c| = |(a, b, c)|. (1.19)
1.3.4 Exercises
1) Let M and N be the midpoints of two opposite sides of a quadrilateralABCD and let P be the midpoint of [MN ]. Prove that
PA+ PB + PC + PD = 0.
2) In a circle of center O, let M be the intersection point of two perpendi-cular chords [AB] and [CD]. Show that
OA+OB +OC +OD = 2OM.
40
3) Consider, in the 3-dimensional space, the parallelograms A1A2A3A4 andB1B2B3B4. Prove that the midpoints of the segments [A1B1], [A2B2],[A3B3] and [A4B4] are the vertices of a new parallelogram.
4) Let ABC be a triangle, H its orthocenter, O the circumcenter (the centerof its circumscribed circle), G the center of gravity of the triangle andA′ diametrically opposed to A (in the circumscribed circle). Then :
a) OA+OB +OC = OH (Sylvester’s formula)
b) HB +HC = HA′
c) HA+HB +HC = 2HO
d) HA+HB +HC = 3HG
e) the points H, G, O are collinear and 2GO = HG (the Euler’sstraight line of the triangle).
5) Let ABC be a triangle and a, b, c the lengths of its sides, respectively.If A1 is the intersection point of the internal bisector of the angle A andBC and M is an arbitrary point, then
MA1 =b
b+ cMB +
c
b+ cMC.
6) The midpoints of the diagonals of a complete quadrilateral are collinear(the Newton-Gauss’ straight line).
7) (Cauchy-Buniakovski-Schwarz) If a1, a2, a3, b1, b2, b3 ∈ R, then
(a1b1 + a2b2 + a2c2)2 ≤ (a21 + a2
2 + a23)(b
21 + b22 + b23).
8) (Cosine theorem in the space) For a tetrahedron ABCD,
cos( AB,CD) =AD2 +BC2 −AC2 −BD2
2AB · CD.
9) (Median line theorem in the space) Let ABCD be a tetrahedron andGA the center of gravity of the BCD side. Then the following equalityholds:
9AG2A = 3(AB2 +AC2 +AD2)− (BC2 + CD2 +BD2).
41
10) Let ∆ABC and ∆A′B′C ′ be two triangles in the same plane, so thatthe perpendicular lines through A,B,C on B′C ′, C ′A′ respectively A′B′
are concurrent. Then so are the perpendicular lines through A′, B′, C ′
on BC, CA respectively AB.
11) Find a vector orthogonal on both u and v.
a) u = −7i+ 3j + k, v = 2i+ 4k
b) u = (−1,−1,−1), v = (2, 0, 2).
12) Let a, b and c be three noncollinear vectors. Show that there exists atriangle ABC with BC = a, CA = b and AB = c if and only if
a× b = b× c = c× a.
13) Find the area of the triangle having the vertices A(1, 0, 1), B(0, 2, 3),and C(2, 1, 0).
14) Prove that:
a) a× (b× c) = (a · c) · b− (a · b) · c =∣∣∣∣ b c
a · b a · c
∣∣∣∣b) (a× b)× c = (a · c) · b− (b · c) · a =
∣∣∣∣ b a
b · c a · c
∣∣∣∣15) Verify the Laplace’s formula:
(a× b) · (c× d) =∣∣∣∣a · c a · db · c b · d
∣∣∣∣ .16) Prove that
(a× b)× (c× d) = (a, c, d) · b− (b, c, d) · a = (a, b, d) · c− (a, b, c) · d.
17) Show that(u× v, v × w,w × u) = (u, v, w)2.
18) The mutual vectors of the vectors u, v, w, supposed to be not in thesame plane, are
u′ =v × w
(u, v, w), v′ =
w × u
(u, v, w), w′ =
u× v
(u, v, w).
42
a) Find the mutual vectors of i, j, k.
b) If a = xu+ yv + zw, prove that
x = a · u′, y = a · v′, z = a · w′.
c) Show that the mutual vectors of u′, v′, w′ are respectively u, v, w.
19) Let a, b and c be the lengths of the sides of the triangle ∆ABC and Rthe radius of its circumscribed circle. Then:
a) OH2 = 9R2 − (a2 + b2 + c2);
b) OG2 = R2 − 19(a2 + b2 + c2).
20) If E and F are the midpoints of the diagonals AC and BD of the convexequadrilateral ABCD, then
AB2 +BC2 + CD2 +DA2 = AC2 +BD2 + 4EF 2.
21) Let K, L, M and N be the midpoints of the sides AB, BC, CD andrespectively DA of the convexe quadrilateral ABCD. Then
AB2 + CD2 −AD2 −BC2 = 2(LN2 −KM2) = 2AC ·BD cosϕ,
where ϕ = (AC,BD).
21) Let ABC be an arbitrary triangle, G its center of gravity, I its incenter,H its orthocenter and O the center of its circumscribed circle. If P isan arbitrary point and rA = PA, rB = PB, rC = PC, then:
a) PG =rA + rB + rC
3;
b) PI =a · rA + b · rB + c · rC
a+ b+ c;
c) PH =tanA · rA + tanB · rB + tanC · rC
tanA+ tanB + tanC;
d) PO =sin 2A · rA + sin 2B · rB + sin 2C · rC
sin 2A+ sin 2B + sin 2C.
22) Let d be a line in the plane of the triangle ∆ABC and A′, B′ and C ′ theorthogonal projections of the vertices A, B, respectively C on d. Provethat the orthogonal lines through A′, B′ respectively C ′ on BC, CArespectively AB are concurrent.
43
23) The radius of the sphere tangent at the point O to the face (BOC) ofthe tetrahedron OABC and passing through the point A is
R =a sinα
2(1 + 2 cosα cosβ cos γ − cos2 α− cos2 β − cos2 γ)12
,
where a = OA, α = BOC, β = COA, γ = AOB.
44
Chapter 2
Two-Dimensional AnalyticGeometry
2.1 Several Equations of Lines
Analytic Geometry allows us to express in an algebraic language geometricproperties of objects. We start here with the Euclidean plane, whose basicelements are the points and the lines. We associate a Cartesian system ofcoordinates, in which an arbitrary point is characterized by two real numbers,its Cartesian coordinates, as we saw in the first chapter. We shall presentdifferent equations of lines in a 2-dimensional Euclidean space.
2.1.1 Parametric Equations of Lines
A line d can be determined by specifying a point P0(x0, y0) on the line anda nonzero vector v(a, b), parallel to the line (the direction of the line).
Figure 2.1:
45
The line d in a 2-space, passing through the point P0(x0, y0) and parallelto the nonzero vector v(a, b) has the parametric equations
d :{x = x0 + aty = y0 + bt
t ∈ R. (2.1)
Indeed, for any point P (x, y) on the line d, the vectors P0P and v are linearlydependent, since they are parallel, hence there exists t ∈ R such that P0P = tv.Identifying the components of these two vectors, respectively, one obtains
d :{x− x0 = aty − y0 = bt
which leads to the formula (2.1).The vector v is said to be the director vector of the line d.
2.1.2 Vector Equations of Lines
The vector language can be used to express the parametric equations of theline (2.1) in a shorter form. The line d is again the line passing through P0
and parallel to the vector v. Choosing an arbitrary point O in the plane, onecan characterize any point P by its position vector, i.e. the vector having theoriginal point O and the terminal point P .
Figure 2.2:
As before, the point P belongs to the line d if and only if the vectors P0Pand v are linearly dependent. This means that there exists t ∈ R, such thatP0P = tv. But P0P = OP −OP0 = r− r0 (see Figure 3.4), hence tv = r− r0,and the vector equation of the line passing through P0 and of director vectorv is
r = r0 + tv. (2.2)
46
2.1.3 Symmetric Equations of Lines
If, in (2.1), one expresses twice the parameter t, one obtains the symmetricequation of the line d passing through the point P0(x0, y0) and of directorvector v(a, b):
x− x0
a=y − y0
b. (2.3)
Remark : The vector v is a nonzero vector, so that at least one of thedenominators a and b is different from zero. Suppose that a = 0. Then, (2.1)
becomes{
x = x0
y = y0 + bt. In fact,
x− x0
0=y − y0
bis just a convenient way to
write that the numerator is zero when the denominator is zero. This is theparticular case of a line which is parallel to Oy. If b = 0, one obtains theequations of a line parallel to Ox.
2.1.4 General Equations of Lines
A simple computation shows that (2.3) can be written in the form
Ax+By + C = 0, with A2 +B2 6= 0, (2.4)
meaning that any line from the 2-space is characterized by a first degree equa-tion. Conversely, such of an equation represents a line, since the formula (2.4)
is equivalent tox+ C
A
−BA
=y
1and this is the symmetric equation of the line
passing through P0
(−C
A, 0
)and parallel to v
(−B
A, 1
).
The equation (2.4) is called general equation of the line.
2.1.5 Reduced Equations of Lines
Consider a line given by its general equation Ax + By + C = 0, where atleast one of the coefficients A and B is nonzero. One may suppose that B 6= 0,so that the equation can be divided by B. One obtains
y = mx+ n (2.5)
which is said to be the reduced equation of the line.
Remark : If B = 0, (2.4) becomes Ax+ C = 0, or x = −C
A, a line parallel
to Oy. (In the same way, if A = 0, one obtains the equation of a line parallelto Ox).
47
Let d be a line of equation y = mx+n in a Cartesian system of coordinatesand suppose that the line is not parallel to Oy. Let P1(x1, y1) and P2(x2, y2)be two different points on d and ϕ be the angle determined by d and Ox (see
Figure 2.3); ϕ ∈ [0, π] \ {π
2}.
Figure 2.3:
The points P1(x1, y1) and P2(x2, y2) belong to d, hence{y1 = mx1 + ny2 = mx2 + n
,
and x2 6= x1, since d is not parallel to Oy. Then,
m =y2 − y1
x2 − x1= tanϕ. (2.6)
The number m = tanϕ is called the angular coefficient of the line d.It is immediate that the equation of the line passing through the point
P0(x0, y0) and of the given angular coefficient m is
y − y0 = m(x− x0). (2.7)
2.1.6 Equations of Lines Determined by Two Points
A line can be uniquely determined by two distinct points P1(x1, y1) andP2(x2, y2) on the line. The line can be seen to be the line passing through thepoint P1(x1, y1) and having P1P2(x2−x1, y2−y1) as director vector, thereforeits equation is
d :x− x1
x2 − x1=
y − y1
y2 − y1. (2.8)
48
The equation (2.8) can be put in the form∣∣∣∣∣∣x y 1x1 y1 1x2 y2 1
∣∣∣∣∣∣ = 0. (2.9)
Given three points P1(x1, y1), P2(x2, y2) and P3(x3, y3), they are collinear ifand only if ∣∣∣∣∣∣
x1 y1 1x2 y2 1x3 y3 1
∣∣∣∣∣∣ = 0.
2.1.7 Exercises
1) The sides [BC], [CA], [AB] of the triangle ∆ABC are divided by thepoints M , N respectively P into the same ratio k. Prove that the trian-gles ∆ABC and ∆MNP have the same center of gravity.
2) Sketch the graph of x2 − 4xy + 3y2 = 0.
3) Find the equation of the line passing through the intersection point ofthe lines
d1 : 2x− 5y − 1 = 0, d2 : x+ 4y − 7 = 0
and through a point M which divides the segment [AB], A(4,−3),
B(−1, 2), into the ratio k =23.
4) Let A be a mobile point on the Ox axis and B a mobile point on Oy, so
that1OA
+1OB
= k (constant). Prove that the lines AB pass through afixed point.
5) Find the equation of the line passing through the intersection point of
d1 : 3x− 2y + 5 = 0, d2 : 4x+ 3y − 1 = 0
and crossing the Oy axis at the point A with OA = 3.
6) Find the parametric equations of the line through P1 and P2, when
a) P1(3,−2), P2(5, 1);
b) P1(4, 1), P2(4, 3).
49
7) Find the parametric equations of the line through P (−5, 2) and parallelto v(2, 3).
8) Show that the equations
x = 3− t, y = 1 + 2t and x = −1 + 3t, y = 9− 6t
represent the same line.
9) Find the vector equation of the line passing through P1 and P2, when
a) P1(2,−1), P2(−5, 3);
b) P1(0, 3), P2(4, 3).
2.2 Parallelism and Orthogonality
2.2.1 Intersection of Two Lines
Let d1 : a1x + b1y + c1 = 0 and d2 : a2x + b2y + c2 = 0 be two lines in E2.The solution of the system of equation{
a1x+ b1y + c1 = 0a2x+ b2y + c2 = 0
will give the set of the intersection points of d1 and d2.
1) Ifa1
a26=b1
b2, the system has a unique solution (x0, y0) and the lines have
a unique intersection point P0(x0, y0). They are secant.
2) Ifa1
a2=b1
b26=c1
c2, the system is not compatible, and the lines have no
points in common. They are parallel.
3) Ifa1
a2=b1
b2=c1
c2, the system has an infinity of solutions, and the lines
coincide. They are identical.
If di : aix+ biy+ ci = 0, i = 1, 3 are three lines in E2, then they are concurrentif and only if ∣∣∣∣∣∣
a1 b1 c1a2 b2 c2a3 b3 c3
∣∣∣∣∣∣ = 0. (2.10)
50
2.2.2 Bundle of Lines
The set of all the lines passing through a given point P0 is said to be abundle of lines. The point P0 is called the vertex of the bundle.
If the point P0 is of coordinates P0(x0, y0), then the equation of the bundleof vertex P0 is
r(x− x0) + s(y − y0) = 0, (r, s) ∈ R2 \ {(0, 0)}. (2.11)
Figure 2.4:
Remark : One may suppose that s 6= 0 and divide in (2.11) by s. Oneobtains the reduced equation of the bundle,
y − y0 = m(x− x0), m ∈ R, (2.12)
in which the line x = x0 is missing. Analogously, if r 6= 0, one obtains thebundle, except the line y = y0.
If the point P0 is given as the intersection of two lines, then its coordinatesare the solution of the system{
d1 : a1x+ b1y + c1 = 0d2 : a2x+ b2y + c2 = 0
,
supposed to be compatible. The equation of the bundle of lines through P0 is
r(a1x+ b1y + c1) + s(a2x+ b2y + c2) = 0, (r, s) ∈ R2 \ {(0, 0)}. (2.13)
Remark : As before, if r 6= 0 (or s 6= 0), one obtains the reduced equationof the bundle, containing all the lines through P0, except d1 (respectively d2).
51
2.2.3 The Angle of Two Lines
Let d1 and d2 be two concurrent lines, given by their reduced equations:
d1 : y = m1x+ n1 and d2 : y = m2x+ n2.
The angular coefficients of d1 and d2 are m1 = tanϕ1 and m2 = tanϕ2 (see
Figure 2.5). One may suppose that ϕ1 6=π
2, ϕ2 6=
π
2, ϕ2 ≥ ϕ1, such that
ϕ = ϕ2 − ϕ1 ∈ [0, π] \ {π
2}.
Figure 2.5:
The angle determined by d1 and d2 is given by
tanϕ = tan(ϕ2 − ϕ1) =tanϕ2 − tanϕ1
1 + tanϕ1 tanϕ2,
hencetanϕ =
m2 −m2
1 +m1m2. (2.14)
1) The lines d1 and d2 are parallel if and only if tanϕ = 0, therefore
d1 ‖ d2 ⇐⇒ m1 = m2. (2.15)
2) The lines d1 and d2 are orthogonal if and only if they determine an angle
ofπ
2, hence
d1⊥d2 ⇐⇒ m1m2 + 1 = 0. (2.16)
52
2.2.4 Exercises
1) Given the line d : 2x+ 3y + 4 = 0, find the equation of a line d1 passingthrough the point M0(2, 1), in the following situations:
a) d1 is parallel with d;
b) d1 is orthogonal on d;
c) the angle determined by d and d1 is ϕ =π
4.
2) The vertices of the triangle ∆ABC are the intersection points of thelines
d1 : 4x+ 3y − 5 = 0, d2 : x− 3y + 10 = 0, d3 : x− 2 = 0.
1. Find the coordinates of A, B, C.
2. Find the equations of the median lines of the triangle.
3. Find the equations of the heights of the triangle.
3) Find the coordinates of the symmetrical of the point P (−5, 13) withrespect to the line d : 2x− 3y − 3 = 0.
4) Find the coordinates of the point P on the line d : 2x − y − 5 = 0 forwhich the sum AP + PB is minimum, when A(−7, 1) and B(−5, 5).
5) Find the coordinates of the circumcenter (the center of the circumscribedcircle) of the triangle determined by the lines 4x−y+2 = 0, x−4y−8 = 0and x+ 4y − 8 = 0.
6) Prove that, in any triangle ∆ABC, the orthocenter H, the center ofgravity G and the circumcenter O are collinear.
7) Given the bundle of lines of equations (1 − t)x + (2 − t)y + t − 3 = 0,t ∈ R and x+ y − 1 = 0, find:
a) the coordinates of the vertex of the bundle;
b) the equation of the line in the bundle which cuts Ox and Oy in Mrespectively N , such that OM2 ·ON2 = 4(OM2 +ON2).
8) Let B be the bundle of vertex M0(5, 0). An arbitrary line from B in-tersects the lines d1 : y − 2 = 0 and d2 : y − 3 = 0 in M1 respectivelyM2. Prove that the line passing through M1 and parallel to OM2 passesthrough a fixed point.
53
9) The vertices of the quadrilateralABCD areA(4, 3), B(5,−4), C(−1,−3)and D((−3,−1).
a) Find the coordinates of the points {E} = AB ∩ CD and{F} = BC ∩AD;
b) Prove that the midpoints of the segments [AC], [BD] and [EF ] arecollinear.
10) Let M be a point whose coordinates satisfy
4x+ 2y + 83x− y + 1
=52.
a) Prove that M belongs to a fixed line;
b) Find the minimum of x2 + y2, when M ∈ d \ {M0(−1,−2)}.
11) Find the geometric locus of the points whose distances to two orthogonallines have a constant ratio.
54
Chapter 3
Three-Dimensional AnalyticGeometry
In a 3-dimensional Euclidean space E3, endowed with a rectangular systemof coordinates Oxyz, a point P ∈ E3 is characterized by three real numbers,the coordinates of the point, P (x, y, z). We study in this chapter the planesand the lines in the 3-dimensional space.
3.1 Analytic Representation of Planes
A plane π in the 3-dimensional space is uniquely determined by specifyinga point P0(x0, y0, z0) in the plane and a nonzero vector n(a, b, c), orthogonalto the plane. n is called the normal vector to the plane π.
Figure 3.1:
An arbitrary point P (x, y, z) is contained into the plane π if and only if
55
n⊥P0P , or n · P0P = 0. But P0P (x − x0, y − y0, z − z0) and one obtainsthe normal equation of the plane π containing the point P0(x0, y0, z0) and ofnormal vector n(a, b, c).
a(x− x0) + b(y − y0) + c(z − z0) = 0. (3.1)
Remark : The equation (3.1) can be written in the form ax+by+cz+d = 0.
Theorem 3.1.1. Given a, b, c, d ∈ R, with a2 + b2 + c2 > 0, the equation
ax+ by + cz + d = 0 (3.2)
represents a plane, having n(a, b, c) as normal vector.
Proof: One may suppose that a 6= 0. The equation (3.2) can be put in
the form a
(x+
d
a
)+ by + cz = 0, which represents the plane containing the
point P
(−d
a, 0, 0
)and having n(a, b, c) as normal vector. �
The equation (3.2) is called the general equation of the plane.Given a fixed point O in the 3-space, any point P is characterized by its
position vector rP = OP .
Theorem 3.1.2. a) The vector equation of the plane π, determined by threenoncollinear points A, B and C, is
r = (1− α− β)rA + αrB + βrC , α, β ∈ R. (3.3)
b) The vector equation of the plane π, determined by a point A and two non-parallel directions v1 and v2 contained into the plane, is
r = rA + αv1 + βv2, α, β ∈ R. (3.4)
Proof: a) For any arbitrary point M ∈ π, the vectors AB, AC and AMare linearly dependent, since they are coplanar (see Figure 3.2 a), i.e. thereexist the scalars α and β ∈ R, such that
AM = αAB + βAC.
By expressing each vector in this equality, one gets
rM − rA = α(rB − rA) + β(rC − rA),
56
Figure 3.2:
or, equivalently,rM = (1− α− β)rA + αrB + βrC .
b) Similarly, for any point M ∈ π, the vectors AM , v1 and v2 are linearlydependent (see Figure 3.2 b), such that there exist α, β ∈ R, with
AM = αv1 + βv2,
orrM − rA = αv1 + βv2
andrM = rA + αv1 + βv2. �
If the points A, B and C which determine the plane π are of coordinatesA(xA, yA, zA), B(xB, yB, zB) and C(xC , yC , zC) and an arbitrary point of π isM(x, y, z), then the equation (3.3) decomposes into three linear equations:
x = (1− α− β)xA + αxB + βxC
y = (1− α− β)yA + αyB + βyC
z = (1− α− β)zA + αzB + βzC
.
This system must have an infinity of solutions (α, β), so that∣∣∣∣∣∣∣∣x y z 1xA yA zA 1xB yB zB 1xC yC zC 1
∣∣∣∣∣∣∣∣ = 0, (3.5)
which is the analytic equation of the plane determined by three noncollinearpoints.
57
The points A, B, C and D are coplanar if and only if:∣∣∣∣∣∣∣∣xA yA zA 1xB yB zB 1xC yC zC 1xD yD zD 1
∣∣∣∣∣∣∣∣ = 0. (3.6)
Replacing now, in (3.4), the vectors v1(p1, q1, r1) and v2(p2, q2, r2) and thepoints A(xA, yA, zA) and M(x, y, z), the equation (3.4) becomes
x = xA + αp1 + βp2
y = yA + αq1 + βq2z = zA + αr1 + βr2
, α, β ∈ R, (3.7)
and these are the parametric equations of the plane. Again, this system musthave an infinity of solutions (α, β), so that∣∣∣∣∣∣
x− xA y − yA z − zAp1 q1 r1p2 q2 r2
∣∣∣∣∣∣ = 0, (3.8)
which is the analytic equation of the plane determined by a point and twononparallel directions.
3.1.1 Exercises
1) Let P1(x1, y1, z1) and P2(x2, y2, z2) be two different points. Prove thatthe equations of the plane containing P1 and P2 and parallel to the vectora(l,m, n) is ∣∣∣∣∣∣
x− x1 y − y1 z − z1x2 − x1 y2 − y1 z2 − z1
l m n
∣∣∣∣∣∣ = 0.
2) Find the equation of the plane containing P (2, 1,−1) and perpendicularon the vector n(1,−2, 3).
3) Find the equation of the plane determined by O(0, 0, 0), P1(3,−1, 2) andP2(4,−2,−1).
4) Find the equation of the plane containing P (3, 4,−5) and parallel toboth a1(1,−2, 4) and a2(2, 1, 1).
5) Find the equation of the plane containing the points P1(2,−1,−3) andP2(3, 1, 2) and parallel to the vector a(3,−1,−4).
58
6) Find the equation of the plane passing through P (7,−5, 1) and whichdetermines on the positive half-axes three segments of the same length.
7) Find the equation of the plane containing the perpendicular lines throughP (−2, 3, 5) on the planes
π1 : 4x+ y − 3z + 13 = 0, π2 : x− 2y + z − 11 = 0.
8) Find the equation of the plane passing through P and having the normalvector n:
a) P (2, 6, 1), n(1, 4, 2);
b) P (1, 0, 0), n(0, 1, 1).
9) Find the equation of the plane passing through the given points:
a) (−2, 1, 1), (0, 2, 3) and (1, 0,−1);
b) (3, 2, 1), (2, 1,−1) and (−1, 3, 2).
10) Show that the points (1, 0,−1), (0, 2, 3), (−2, 1, 1) and (4, 2, 3) are copla-nar.
3.2 Analytic Representations of Lines
As in the case of E2, a line d in the 3-space is completely determined by apoint P0(x0, y0, z0) of the line and a nonzero vector v(p, q, r), parallel to d. Inorder to represent a line in the 3-dimensional case, the parametric equationsare, generally, the most convenient.
If P (x, y, z) is an arbitrary point of the line d, then the vectors P0P and vare linearly dependent in V3 (see Figure 3.3) and there exists t ∈ R, such that
P0P = tv. (3.9)
Since P0P (x − x0, y − y0, z − z0), by decomposing (3.9) in components, oneobtains the parametric equations of the line passing through P0(x0, y0, z0) andparallel to v(a, b, c):
x = x0 + pty = y0 + qtz = z0 + rt
, t ∈ R. (3.10)
The vector v(p, q, r) is called the director vector of the line d.
59
Figure 3.3:
For a fixed point O in the space, the vector P0P can be expressed as thedifference r − r0 (see Figure 3.4) and the equation (3.9) becomes
r = r0 + tv, t ∈ R, (3.11)
said to be the vector equation of the line in 3-space.
Figure 3.4:
Expressing t three times in (3.10), one obtains the symmetric equations ofthe line d:
x− x0
p=y − y0
q=z − z0r
. (3.12)
Remark : The director vector v is a nonzero vector, i.e. at least one of itscomponents is different from zero. As in the 2-dimensional case, if p = 0, for
60
instance, the meaning ofx− x0
0is that x = x0.
A line d can be determined by two different points P1(x1, y1, z1) andP2(x2, y2, z2) of it. In this case, the director vector of d is
P1P2(x2 − x1, y2 − y1, z2 − z1)
and the equations of the line determined by two points are
x− x1
x2 − x1=
y − y1
y2 − y1=
z − z1z2 − z1
. (3.13)
Given two distinct and nonparallel planes π1 : A1x+B1y +C1z +D1 = 0and π2 : A2x + B2y + C2z +D2 = 0 (the planes π1 and π2 are parallel whentheir normal vectors n1(A1, B1, C1) and n2(A2, B2, C2) are parallel, i.e. the
rank of the matrix(A1 B1 C1
A2 B2 C2
)is 1), they have an entire line d in common.
Then, a line in 3-space can be determined as the intersection of two nonparallelplanes:
d :{A1x+B1y + C1z +D1 = 0A2x+B2y + C2z +D2 = 0
, with rank(A1 B1 C1
A2 B2 C2
)= 2.
(3.14)
3.2.1 Exercises
1) Let d1 and d2 be two lines in E3, given by
d1 :x− 1
2=y + 1−1
=z − 5
6and d2 :
x− 11
=y + 1
1=z − 5−3
.
b) Find the parametric equations of d1 and d2.
b) Prove that they are incident and find the coordinates of their in-tersection point.
c) Find the equation of the plane determined by d1 and d2.
2) Given two lines
d1 : x = 1 + t, y = 1 + 2t, z = 3 + t, t ∈ R
andd2 : x = 3 + s, y = 2s, z = −2 + s, s ∈ R,
prove that d1 ‖ d2 and find the equation of the plane determined by d1
and d2.
61
3) Find the parametric equations of the line{−2x+ 3y + 7z + 2 = 0x+ 2y − 3z + 5 = 0
.
4) Find the parametric equations of the line passing through P1 and P2:
a) P1(5,−2, 1) and P2(2, 4, 2);
b) P1(4, 0, 7) and P2(−1,−1, 2)
5) Find the parametric equations of the line passing through (−1, 2, 4) andparallel to v(3,−4, 1).
6) Find the equations of the line passing through the origin and parallel tothe line
x = ty = −1 + tz = 2
.
3.3 Relative Positions for Lines and Planes
3.3.1 Relative Positions of Two Lines
Let d1 and d2 be two lines in E3, of director vectors v1(p1, q1, r1) 6= 0,respectively v2(p2, q2, r2) 6= 0. The parametric equations of these lines are
d1 :
x = x1 + p1ty = y1 + q1tz = z1 + r1t
, t ∈ R; and d2 :
x = x2 + p2sy = y2 + q2sz = z2 + r2s
, s ∈ R.
The set of the intersection points of d1 and d2 is given by the set of thesolutions of the system of equations
x1 + p1t = x2 + p2sy1 + q1t = y2 + q2sz1 + r1t = z2 + r2s
. (3.15)
• If the system (3.15) has a unique solution (t0, s0), then the lines d1 andd2 have exactly one intersection point P0, corresponding to t0 (or s0).One says that the lines are concurrent (or incident); {P0} = d1 ∩ d2.The vectors v1 and v2 are, necessarily, linearly independent.
62
Figure 3.5:
• If the system (3.15) has an infinity of solutions, then the two lines havean infinity of points in common, so that the lines coincide. They areidentical ; d1 = d2. There exists α ∈ R∗ such that v1 = αv2 (theirdirector vectors are linearly dependent) and any arbitrary point of d1
belongs to d2 (and reciprocally).
• Suppose that the above system of equations is not compatible. Thereare two possible situations.
– If the director vectors are linearly dependent (there exists α ∈ R∗
such that v1 = αv2 or, equivalently,p1
p2=q1
q2=r1
r2), then the lines
are parallel ; d1 ‖ d2.
Figure 3.6:
– If the director vectors are linearly independent, then one deals withskew lines (nonparallel and nonincident); d1 ∩ d2 = ∅ and d1 ∦ d2.
3.3.2 Relative Positions of Two Planes
Letπ1 : a1x+ b1y + c1z + d1 = 0, n1(a1, b1, c1) 6= 0
63
andπ2 : a2x+ b2y + c2z + d2 = 0, n2(a2, b2, c2) 6= 0
be two planes, having the normal vectors n1, respectively n2.The intersection of these planes is given by the solution of the system of
equations {π1 : a1x+ b1y + c1z + d1 = 0π2 : a2x+ b2y + c2z + d2 = 0
. (3.16)
• If rank(a1 b1 c1a2 b2 c2
)= 2, then the system (3.16) is compatible and the
planes have a line in common. They are incident ; π1 ∩ π2 = d.
Figure 3.7:
• If rank(a1 b1 c1a2 b2 c2
)= 1, then the rows of the matrix are proportional,
a1
a2=b1
b2=c1
c2, which means that the normal vectors of the planes are
linearly dependent. There are two possible situations:
Figure 3.8:
64
– Ifa1
a2=b1
b2=c1
c26=d1
d2, then the system (3.16) is not compatible,
and the planes are parallel ; π1 ‖ π1.
– Ifa1
a2=b1
b2=c1
c2=d1
d2, then the planes are identical ; π1 = π2.
Bundle of Planes
Given a line d, the set of all the planes containing the line d is said to bethe bundle of planes through d. Let us suppose that d is determined as theintersection of two planes π1 and π2, i.e.
d :{π1 : a1x+ b1y + c1z + d1 = 0π2 : a2x+ b2y + c2z + d2 = 0
, with rank(a1 b1 c1a2 b2 c2
)= 2.
The equation of the bundle is
λ1π1 + λ2π2 = 0, (λ1, λ2) ∈ R2 \ {(0, 0)}. (3.17)
Remark : Since not both λ1 and λ2 are zero, one may suppose that λ1 6= 0and divide in (3.17) by λ1; one obtains the reduced equation of the bundle:
π1 + λπ2 = 0,
which contains all the planes through d, except π2.
3.3.3 Relative Positions of a Line and a Plane
Let
d :
x = x1 + pty = y1 + qtz = z1 + rt
, p2 + q2 + r2 > 0
be a line of director vector v(p, q, r) and
π : ax+ by + cz + d = 0, a2 + b2 + c2 > 0
be a plane of normal vector n(a, b, c).The intersection between d and π is given by the solutions of the equation
a(x1 + pt) + b(y1 + qt) + c(z1 + rt) + d = 0. (3.18)
• If (3.18) has a unique solution t0, then d and π have one intersectionpoint P0, corresponding to the parameter t0. The line and the plane areincident ; d ∩ π = {P0}.
65
Figure 3.9:
• If (3.18) has an infinity of solutions, then d and π have the entire line din common and d is contained into π; d ⊂ π. In this case, the normalvector n of π is orthogonal on the director vector v of d (then n · v = 0,or ap+ bq + cr = 0) and any point of d is contained into π.
Figure 3.10:
• If (3.18) has no solutions, then the line d is parallel to the plane π; d ‖ π.
Figure 3.11:
3.3.4 Exercises
1) Find the equations of the line passing through P (6, 4,−2) and parallel
to the line d :x
2=y − 1−3
=z − 5
6.
66
2) Given the lines
d1 : x = 4− 2t, y = 1 + 2t, z = 9 + 3t
and
d2 :x− 1
2=y + 2
3=z − 4
2,
find the intersection points between the two lines and the coordinateplanes.
3) Let d1 and d2 be the lines given by
d1 : x = 3 + t, y = −2 + t, z = 9 + t, t ∈ R
andd2 : x = 1− 2s, y = 5 + s, z = −2− 5s, s ∈ R.
a) Prove they are coplanar.
b) Find the equation of the line passing through the point P (4, 1, 6)and orthogonal on the plane determined by d1 and d2.
4) Prove that the intersection lines of the planes π1 : 2x− y + 3z − 5 = 0,π2 : 3x+ y + 2z − 1 = 0 and π3 : 4x+ 3y + z + 2 = 0 are parallel.
5) Verify that the lines
d1 :x− 3
1=y − 8
3=z − 3
4
and
d2 :x− 4
1=y − 9
2=z − 9
5are coplanar and find the equation of the plane determined by the twolines.
6) Determine whether the line x = 3 + 8ty = 4 + 5tz = −3− t
is parallel to the plane x− 3y + 5z − 12 = 0.
67
7) Find the intersection point between the linex = 3 + 8ty = 4 + 5tz = −3− t
and the plane x− 3y + 5z − 12 = 0.
8) Prove that the lines
d1 :
x = 1 + 4ty = 5− 4tz = −1 + 5t
and d2 :
x = 2 + 8ty = 4− 3tz = 5 + t
are skew.
9) Find the parametric equations of the line passing through (5, 0,−2) andparallel to the planes x− 4y + 2z = 0 and 2x+ 3y − z + 1 = 0.
10) Find the equation of the plane containing the point P (2, 0, 3) and theline
d :
x = −1 + ty = tz = −4 + 2t
11) Show that the line x = 0y = tz = t
is contained into the plane 6x+ 4y − 4z = 0.
12) Let M1(2, 1,−1) and M2(−3, 0, 2) be two points. Find:
a) the equation of the bundle of planes passing through M1 and M2;
b) the plane π from the bundle, which is orthogonal on xOy;
c) the plane from the bundle, which is orthogonal on π.
13) Given the points A(1, 2α, α), B(3, 2, 1), C(−α, 0, α) and D((−1, 3,−3),find the parameter α, such that the bundle of planes passing throughAB has a common point with the bundle of planes passing through CD.
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14) Given the planesπ1 : 2x+ y − 3z − 5 = 0
andπ2 : x+ 3y + 2z + 1 = 0,
find the equations of the bisector planes of the dihedral angle and choosewhich one belongs to the acute dihedral angle.
3.4 Metric Problems Concerning Angles
3.4.1 The Angle Determined by Two Lines
Let d1 and d2 be two lines on E3, whose director vectors are v1 respectivelyv2. The angle determined by d1 and d2 is considered to be the acute or rightangle formed by d1 and d2. It is denoted by (d1, d2).
Theorem 3.4.1.1. The measure of the angle determined by d1 and d2 is givenby
m(d1, d2) =
{m(v1, v2), if v1 · v2 ≥ 0
π −m(v1, v2), if v1 · v2 < 0(3.19)
Figure 3.12:
Proof: The assertion is immediate. �Using the definition of the dot product of two vectors, (3.19) becomes
m(d1, d2) =
arccos
v1 · v2
|v1||v2|, if v1 · v2 ≥ 0
π − arccosv1 · v2
|v1||v2|, if v1 · v2 < 0
(3.20)
Remark : Two (concurrent or skew) lines d1 and d2, having the directorvectors v1(p1, q1, r1), respectively v2(p2, q2, r2), are orthogonal if their directorvectors are orthogonal.
d1⊥d2 ⇐⇒ v1 · v2 = 0 ⇐⇒ p1p2 + q1q2 + r1r2 = 0. (3.21)
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3.4.2 The Angle Determined by a Line and a Plane
Let d be a line of director vector v(p, q, r) and π be a plane of normal vectorn(a, b, c). The angle determined by d and π, denoted by (d, π), is the angledetermined by d and the orthogonal projection d′ of d on π.
Figure 3.13:
Theorem 3.4.2.1. The measure of the angle determined by the line d and theplane π is given by
m(d, π) =
π
2−m(v, n), if v · n ≥ 0
m(v, n)−π
2, if v · n < 0
(3.22)
Figure 3.14:
The formula (3.22) has the alternative form
m(d, π) =
π
2− arccos
v · n|v||n|
, if v · n ≥ 0
arccosv · n|v||n|
−π
2, if v · n < 0
(3.23)
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Remarks:
1) The line d is parallel to the plane π if the vector v is orthogonal to n,hence
d ‖ π ⇐⇒ v · n = 0 ⇐⇒ pa+ qb+ rc = 0. (3.24)
2) The line d is orthogonal to the plane π if v is parallel to n. Then
d⊥π ⇐⇒ v ‖ n ⇐⇒ ∃ α ∈ R∗ : n = αd. (3.25)
3.4.3 The Angle Determined by Two Planes
Let π1 and π2 be two planes of normal vectors n1(a1, b1, c1), respectivelyn2(a2, b2, c2). The angle determined by π1 and π2, denoted by (π1, π2), is theacute or right dihedral angle of π1 and π2.
Figure 3.15:
Theorem 3.4.3.1. The measure of the angle determined by π1 and π2 is givenby
m(π1, π2) =
{m(n1, n2), if n1 · n2 ≥ o
π −m(n1, n2), if n1 · n2 < o(3.26)
The formula (3.26) can be written in the form
m(π1, π2) =
arccos
n1 · n2
|n1||n2|, if n1 · n2 ≥ o
π − arccosn1 · n2
|n1||n2|, if n1 · n2 < o
(3.27)
Remark : The planes π1 and π2 are orthogonal if and only if their normalvectors are orthogonal, hence
π1⊥π2 ⇐⇒ n1 · n2 = 0 ⇐⇒ a1a2 + b1b2 + c1c2 = 0. (3.28)
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Figure 3.16:
3.4.4 Exercises
1) Find the angle determined by d1 and d2:
a) d1 : x = 4− t, y = 3 + 2t, z = −2t, t ∈ Rd2 : x = 5 + 2s, y = 1 + 3s, z = 5− 6s, s ∈ R.
b) d1 :x− 1
2=y + 5
7=z − 1−1
, d2 :x+ 3−2
=y − 9
1=z
4.
2) Find the angle determined by the planes
π1 : x−√
2y + z − 1 = 0 and π2 : x+√
2y − z + 3 = 0.
3) Find the equations of the projection of the line
d :{
2x− y + z − 1 = 0x+ y − z + 1 = 0
on the plane π : x+ 2y − z = 0.
4) Find the orthogonal projection of the point P (2, 1, 1) on the planeπ : x+ y + 3z + 5 = 0.
5) Find the angle determined by the lines
d1 :{x+ 2y + z − 1 = 0x− 2y + z + 1 = 0
and d2 :{x− y − z − 1 = 0x− y + 2x+ 1 = 0
.
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6) Find the angle determined by the planes
π1 : x+ 3y + 2z + 1 = 0 and π2 : 3x+ 2y − z − 6 = 0.
7) Find the angle determined by the plane xOy and the line M1M2, whereM1(1, 2, 3) and M2(−2, 1, 4).
3.5 Metric Problems Concerning Distances
3.5.1 The Distance From a Point to a Plane
Let P0(x0, y0, z0) be a point and π : ax+by+cz+d = 0 (with a2+b2+c2 > 0)be a plane in E3.
Theorem 3.5.1.1. The distance from the point P0(x0, y0, z0) to the planeπ : ax+ by + cz + d = 0 is given by
d(P0, π) =|ax0 + by0 + cz0 + d|
√a2 + b2 + c2
. (3.29)
Proof: Let P1(x1, y1, z1) be an arbitrary point on the plane π and letn(a, b, c) be the normal vector of π.
Figure 3.17:
The distance from P0 to π is the length of the orthogonal projection of thevector P1P0 on the direction of n (see Figure 3.17). Then, one has
d(P0, π) = |prnP1P0| =|P1P0 · n|
|n|=|a(x0 − x1) + b(y0 − y1) + c(z0 − z1)|√
a2 + b2 + c2=
=|ax0 + by0 + cz0 − (ax1 + by1 + cz1)|√
a2 + b2 + c2=|ax0 + cy0 + cz0 + d|
√a2 + b2 + c2
.�
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3.5.2 The Distance From a Point to a Line
Given a point P0(x0, y0, z0) and a line d :
x = x1 + pty = y1 + qtz = z1 + rt
, t ∈ R, with
p2 + q2 + r2 > 0, we present two ways to find the distance from P0 to d.
I) Let v(p, q, r) be the director vector of d and P1(x1, y1, z2) be an arbitrarypoint on d. The distance from P0 to d is the altitude of the parallelogramdetermined by v and P1P0 (see Figure 3.18).
Figure 3.18:
This altitude can be expressed using the area of the parallelogram andone has
d(P0, d) =|v × P1P0|
|v|. (3.30)
II) Let π be the plane passing through P0 and orthogonal on d. Its equationis
π : p(x− x0) + q(y − y0) + r(z − z0) = 0.
Let P ′0 be the intersection point of π and d; {P ′0} = d ∩ π (see Figure3.19). The coordinates of the point P ′0 correspond to the parameter t0,solution of the equation
p(x1 + pt− x0) + q(y1 + qt− y0) + r(z1 + rt− z0) = 0.
Finally, d(P0, d) = d(P0, P′0).
3.5.3 The Distance Between Two Parallel Planes
Let π1 and π2 be two parallel planes. Choose an arbitrary point P1 ∈ π1.Then
d(π1, π2) = d(P1, π2).
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Figure 3.19:
Figure 3.20:
3.5.4 The Distance Between Two Lines
Let d1 and d2 be two lines in the 3-space.
• If the lines are identical or concurrent, then d(d1, d2) = 0.
• If the lines are parallel, it is enough to choose an arbitrary point P1 ∈ d1
and d(d1, d2) = d(P1, d2).
Figure 3.21:
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• If d1 and d2 are skew, there exists a unique line which is orthogonal onboth d1 and d2 and intersects both d1 and d2. The length of the segmentdetermined by these intersection points is the distance between the skewlines.
Suppose that
d1 :
x = x1 + p1ty = y1 + q1tz = z1 + r1t
, t ∈ R and d2 :
x = x2 + p2sy = y2 + q2sz = z2 + r2s
, s ∈ R
are, respectively, the parametric equations of the lines of director vectorsv1(p1, q1, r1) 6= 0, respectively v2(p2, q2, r2) 6= 0.
One can determine the equations of two parallel planes π1 ‖ π2, suchthat d1 ⊂ π1 and d2 ⊂ π2. The normal vector n of these planes has tobe orthogonal on both v1 and v2, hence n = v1 × v2.
Figure 3.22:
Then n(A,B,C), with A =∣∣∣∣q1 r1q2 r2
∣∣∣∣, B =∣∣∣∣r1 p1
r2 p2
∣∣∣∣ and C =∣∣∣∣p1 q1p2 q2
∣∣∣∣.The equations of the planes π1 and π2 are:
π1 : A(x− x1) +B(y − y1) + C(z − z1) = 0
π2 : A(x− x2) +B(y − y2) + C(z − z2) = 0.
Now, the distance between d1 and d2 is the distance between the parallelplanes π1 and π2; d(d1, d2) = d(π1, π2), and one has:
Theorem 3.5.4.1. The distance between two skew lines d1 and d2 isgiven by
d(d1, d2) =|A(x1 − x2) +B(y1 − y2) + C(z1 − z2)|√
A2 +B2 + C2. (3.31)
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3.5.5 Exercises
1) Find the distance from the point P (1, 2,−1) to the line d : x = y = z.
2) Find the distance from (3, 1,−1) to the plane 22x+ 4y − 20z − 45 = 0.
3) Find the equation of the line passing through P (4, 3, 10) and orthogonalon the line
d :x− 1
2=y − 2
4=z − 3
5,
the distance from P to d and the coordinates of the symmetrical P ′ ofP with respect to the line d.
4) Find the distance between the lines
d1 :x− 1
2=y + 1
3=z
1and d2 :
x+ 13
=y
4=z − 1
3.
5) Find the distance from the point (0, 1, 4) to the plane
π : 3x+ 6y − 2z − 5 = 0.
6) Find the distance between the planes
2x− 3y + 4z − 7 = 0 and 4x− 6y + 8z − 3 = 0.
7) Show that the linex+ 1
1=y − 3
2=
z
−1and the plane 2x−2y−2z+3 = 0
are parallel and find the distance between them.
8) Find the geometric locus of the lines passing through a given point andhaving a constant distance to a given line.
9) Let ABCD be a tetrahedron and d a line which intersects the faces ofthe tetrahedron at A′, B′, C ′ respectively D′. Prove that the midpointsof the segments [AA′], [BB′], [CC ′] and [DD′] are coplanar.
10) Let V ABC be a regular quadrilateral pyramid of vertex V , having thesides AB = a and V A = a
√2. Let M be the midpoint of [V A]. The
plane passing through M and orthogonal on V C determines a section inthe pyramid. Find the perimeter and the area of this section.
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11) Let ABCD be a cube with side of length a, M and P be the midpointsof [BC] respectively [AA′], O the center of the cube, O′ the center ofthe face A′B′C ′D′ and S the midpoint of [OO′]. Find the area of thesection determined by the plane (MPS) in the cube.
12) Let ABCD be a cube with side of length a, M and P be the midpoints of[BC] respectively [AA′] and O′ the center of the face A′B′C ′D′. Find theperimeter and the area of the section determined by the plane (MPO′)in the cube.
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Chapter 4
Plane Isometries
4.1 General Properties
The Euclidean plane E2 can be identified with the metric space (R2, d2),where the metric d2 is the Euclidean metric
d2(A,B) =√
(xB − xA)2 + (yB − yA)2, ∀A(xA, yA), B(xB, yB).
The map f : E2 → E2 is said to be an isometry of the plane E2 if f conservesthe distances, i.e.
|f(A)f(B)| = |AB|, ∀A,B ∈ E2.
(One denotes |AB| = d2(A,B)).
Proposition 4.1.1. The image of a segment through an isometry f : E2 → E2
is a segment of the same length.
Proof: Let A and B be two points on E2. It is enough to prove thatf([AB]) = [f(A)f(B)].
Let M ∈ [AB]. Then |AM |+ |MB| = |AB|. Since f is an isometry,
|f(A)f(M)|+ |f(M)f(B)| = |f(A)f(B)|,
which means that f(M) ∈ [f(A)f(B)]. Then f([AB]) ⊆ [f(A)f(B)]).Conversely, take Y ∈ [f(A)f(B)]. Since
|f(A)Y | ≤ |f(A)f(B)| and |f(A)f(B)| = |AB|,
79
there exists X ∈ [AB] such that |AX| = |f(A)Y |. The points X and Y havethe following properties:{
|AX| = |f(A)f(X)| = |f(A)Y |X ∈ [AB] =⇒ f(X) ∈ [f(A)f(B)]
Then f(X) = Y and Y ∈ f([AB]), such that [f(A)f(B)] ⊆ f([AB]). �
Proposition 4.1.2. Let f : E2 → E2 be an isometry. Then:
1) The image of a half-line is a half-line;
2) The image of a line is a line;
3) If A, B and C are three noncollinear points on E2, then so are theirimages f(A), f(B) and f(C);
4) The image of a triangle ∆ABC is triangle ∆f(A)f(B)f(C), such that
∆ABC ≡ ∆f(A)f(B)f(C);
5) The image of an angle AOB is an angle f(A)f(O)f(B) having the samemeasure;
6) Two orthogonal lines are transformed into two orthogonal lines;
7) Two parallel lines are transformed into two parallel lines.
Proposition 4.1.3. Any isometry f : E2 → E2 is a surjective map.
Proof: Choose three noncollinear points A, B and C and let A′, B′ and C ′
be their images (see Figure 4.1).Let X ′ ∈ E2. Let M ′ ∈ A′B′ and N ′ ∈ A′C ′, such that X ′M ′ ‖ A′C ′
and X ′N ′ ‖ A′B′. Since the image of a line through f is a line, there existM ∈ AB and N ∈ AC, such that f(M) = M ′ and f(N) = N ′. Construct theparallelogram AMXN . Since f(M) = M ′ and f(N) = N ′, then f(MX) =M ′X ′ (the image of the line passing through M and parallel to AC is theline passing through f(M) = M ′ and parallel to A′C ′) and f(NX) = N ′X ′,therefore f(X) = X ′ and f is surjective. �
Denote the set of isometries of the plane by Iso(E2);
Iso(E2) = {f : E2 → E2, f isometry}.
80
Figure 4.1:
Proposition 4.1.4. (Iso (E2), ◦) is a group, called the group of isometries ofthe plane.
Proof: (Iso(E2), ◦) is a subgroup of the group of bijective maps from E2 toE2,
S(E2) = {f : E2 → E2, f bijection}.
Indeed, it have been proved that any isometry is surjective. Now, given twopoints A and B such that f(A) = f(B), it follows that |f(A)f(B)| = 0, so|AB| = 0 and A = B, hence f is injective. Then Iso(E2) ⊂ S(E2).
Moreover, if f and g are isometries, then
|f ◦ g−1(A)f ◦ g−1(B)| = |f(g−1(A))f(g−1(B))| =
= |g−1(A)g−1(B)| = |g(g−1(A))g(g−1(B))| = |AB|,
hence f ◦ g−1 is an isometry. �
• A point A ∈ E2 is a fixed point for the isometry f if f(A) = A;
• A line d ∈ E2 is said to be invariant with respect to f if f(d) = d (obvi-ously, a line whose points are all fixed is invariant, while the converse isnot necessarily true).
Proposition 4.1.5. 1) If A and B are two fixed points for the isometry f ,then any point of the line AB is fixed for f ;
2) If the isometry f has three noncollinear fixed points, then f is the iden-tity of the plane, f = 1E2.
81
Proof: 1) Let C be an arbitrary point of the line AB and suppose, for in-stance, that C ∈ [AB]. Then, its image f(C) ∈ [f(A)f(B)] = [AB]. Moreover,|AC|+ |CB| = |AB|, hence |Af(C)|+ |f(C)B| = |AB|. It follows necessarilythat f(C) = C, therefore C is a fixed point for f .
2) Let A, B and C be three noncollinear fixed points for f and choose anarbitrary point X ∈ E2. Let M ∈ AB and N ∈ CA, such that XM ‖ AC andXN ‖ AB (see Figure 4.2).
Figure 4.2:
The points M and N are fixed for f , since they belong to lines determinedby two fixed points. Then, the line MN has only fixed points. Hence, themidpoint O of the segment [MN ] is a fixed point, so that the line AO containsonly fixed points. Therefore, X is a fixed point for f . �
4.2 Symmetries
Let d be a line in E2. The map sd : E2 → E2, given by
sd(P ) = P ′, whereP ′ is the symmetrical of P with respect to the line d,
is called axial symmetry. The line d is the axis of the symmetry.Let be given a point O in the plane. The map sO : E2 → E2, given by
sO(P ) = P ′, whereP ′ is the symmetrical of P with respect to the point P,
is called central symmetry. The point O is the center of the symmetry.A map f : E2 → E2, having the property that f ◦ f = 1E2 , is an involution.
Theorem 4.2.1. The axial symmetries and the central symmetries are invo-lutive isometries.
82
Figure 4.3:
Figure 4.4:
Proof: Let sO be a central symmetry. It os obvious that sO ◦ sO = 1E2 ,hence sO is an involution. Moreover, sO is a bijection and s−1
O = sO.Let P and Q be two arbitrary points on E2 and let P ′ and Q′ be their
symmetricals with respect to O.
Figure 4.5:
The triangles ∆POQ and ∆P ′OQ′ are congruent, hence |PQ| = |P ′Q′|and sO is an isometry.
Similarly, an axial symmetry sd is an isometry with s−1d = sd and an
involution. �
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Remarks:
1) Any point lying on the axis d of an axial symmetry sd is a fixed pointfor sd;
2) Any line d′ which is orthogonal on the axis d of an axial symmetry sd isinvariant with respect to sd (i.e. sd(d′) = d′);
3) The center O of a central symmetry sO is fixed for sO;
4) Any line passing through the center O of a central symmetry sO is in-variant with respect to sO.
We saw in Theorem 4.2.1 that the symmetries are involutions. The fol-lowing result shows that, excepting the identity of E2, the symmetries are theonly involutive isometries.
Proposition 4.2.2. Let f ∈ Iso (E2) be an involutive isometry. Then f iseither an axial symmetry, or a central symmetry, or 1E2.
Proof: Any involutive isometry has at least a fixed point. Indeed, if A isan arbitrary point on the plane and A′ = f(A), then
f(A′) = f(f(A)) = (f ◦ f)(A) = A.
The image, through f , of the midpoint M of the segment [AA′] is the midpointof the segment [A′A], hence M . Then M is a fixed point for f .
Suppose that f has a unique fixed point O. This point must be the mid-point of any segment [AA′] in the plane, so that for any A ∈ E2, one hasA′ = sO(A). Then, f = sO.
If f has at least two fixed points A and B, then any point of the lined = AB is fixed (see Proposition 4.1.5). If f has another fixed point P , noton d, then f = 1E2 .
Suppose that all the fixed points of f are situated on d. Take an arbitrarypoint M ∈ E2 \ {d} and let M0 be the orthogonal projection of M on d. ThenM0 is a fixed point for f (since it lies on d) and the line MM0 is invariant(since it is orthogonal on d). The image M ′ of M will be on MM0 and|MM0| = |M ′M0| (f is an isometry). Moreover, M ′ 6= M (since M /∈ d, henceM is not a fixed point), and then M ′ = sd(M) and f = sd. �
Theorem 4.2.3. Given A, B, C, A′, B′, C ′ ∈ E2, such that A, B andC are noncollinear and ∆ABC ≡ ∆A′B′C ′, there exists a unique isometryf ∈ Iso(E2), with
f(A) = A′, f(B) = B′, f(C) = C ′.
84
Proof: It is clear that, if ∆ABC 6= ∆A′B′C ′, an isometry having therequired properties does not exists.
We prove first that, if such an isometry does exist, then it is unique. Sup-pose there are two isometries f, g ∈ Iso (E2), such that f(A) = A′, f(B) = B′,f(C) = C ′, and g(A) = A′, g(B) = B′, g(C) = C ′. The map h = f−1 ◦ g isalso an isometry and h(A) = A, h(B) = B, h(C) = C. Hence h has threenoncollinear fixed points, and therefore h = 1E2 and f = g.
Let us prove now the existence of f .
1) If A = A′, B = B′ and C = C ′, one can take f = 1E2 ;
2) If A = A′, B = B′ and C 6= C ′, one can take f = sAB;
3) If A = A′, B 6= B′ and C 6= C ′, let b be the perpendicular line onthe midpoint of the segment [BB′] (which passes through the point A,since |AB| = |AB′|). Then B′ = sb(B). The points A and B are fixedpoints for the isometry sb ◦ f . If (sb ◦ f)(C) = C, then sb ◦ f has threenoncollinear fixed points, sb ◦ f = 1E2 and f = sb. If (sb ◦ f)(C) 6= C,then sb ◦ f = sAB, and f = sb ◦ sAB.
4) If A 6= A′, B 6= B′ and C 6= C ′, let a be the perpendicular line on themidpoint of the segment [AA′]. It is easy to see that (sa ◦ f)(A) = Aand, with the same argument as above, the isometry is given either byf = sa ◦ sb, or by f = sa ◦ sb ◦ sAB. �
Theorem 4.2.4. Given A, B, A′, B′ ∈ E2 (with A and B distinct), such that[AB] ≡ [A′B′], there exist exactly two isometries of E2, with
f(A) = A′ and f(B) = B′.
Proof: Let C be an arbitrary point of the plane E2, not situated on theline AB. There exist exactly two points C ′1 and C ′2 in the plane, such that∆ABC ≡ ∆A′B′C ′1, respectively ∆ABC ≡ ∆A′B′C ′2 (the points C ′1 and C ′2are symmetrical with respect to the line A′B′). Now, Theorem 4.2.3 assuresthe existence of a unique isometry f , such that
f(A) = A′, f(B) = B′, f(C) = C ′1
and a unique isometry g, such that
g(A) = A′, g(B) = B′, g(C) = C ′2. �
Theorem 4.2.5. Any isometry is the product of at the most three axial sym-metries.
85
Proof: Let A, B and C be three noncollinear points, f ∈ Iso (E2) andA′ = f(A), B′ = f(B), C ′ = f(C). Theorem 4.2.3 assures that the isometryf is unique and, following the way in which f was constructed before, it canbe decomposed into at the most three axial symmetries. �
Corollary 4.2.6. The axial symmetries generate the group of isometries.
4.2.1 Exercises
1) Let d be a line and A and B be two points on the plane. Determine theposition of a point M on d, such that the sum |AM |+ |MB| is minimum.
2) Let C be a point inside the angle AOB. Determine two pointsP ∈ (OA and Q ∈ (OB, such that the perimeter of the triangle ∆CPQis minimum.
3) Inscribe in the acute triangle ∆ABC a triangle of minimal perimeter.
4) Let AOB be an angle and O be a point inside the triangle. A variableline d, passing through O, cuts AB at M and BC at N . Prove that thearea of the triangle ∆MBN is minimal if and only if O is the midpointof the segment [MN ].
5) Let ABCD be a parallelogram, ∆ABE and ∆CDF two equilateral tri-angles situated outside the parallelogram, and let G and H be, respec-tively, the centers of two squares constructed on AD, respectively BC,outside the parallelogram. Prove that EGFH is a parallelogram.
6) Let ABCD be a parallelogram and O1, O2 be the incenters of the tri-angles ∆ABC, respectively ∆ADC. Show that AO1CO2 is a parallelo-gram.
7) Let ∆ABC be a triangle and H its orthocenter. Prove that the sym-metricals of H with respect to AB, BC and CA are situated on thecircumscribed circle of the triangle. Also, the symmetricals of H withrespect to the midpoints of [AB], [BC] and [CA] are on the circum-scribed circle.
4.3 Translations
Let v be a vector in V2. The map tv : E2 → E2, given by
tv(M) = M ′, where MM ′ = v,
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is called translation of vector v.
Figure 4.6:
• It is easy to see that t−1v = t−v, therefore
t−1v = tv ⇐⇒ v = 0.
Hence, the unique involutive translation is the identity of the plane(translation of vector 0);
• Any line, having the direction parallel to the direction of v, is invariantwith respect to tv.
Theorem 4.3.1. A product of two central symmetries is a translation. Con-versely, any translation can be decomposed into a product of two central sym-metries.
Proof: Let sA and sB be two central symmetries.
• If A = B, then sA ◦ sB = s2A = 1E2 = t0;
• If A 6= B, let M be an arbitrary point on E2, M ′ = sA(M) and
M ′′ = sB(M ′) = sB(sA(M)) = (sB ◦ sA)(M).
Since [AB] is a midline of the triangle ∆MM ′M ′′, then MM ′′ = 2AB. Inconclusion, sB ◦ sa = t2AB (see Figure 4.7).
Conversely, given the translation tv, one can write tv = sB ◦ sA, where A
and B are two points on the plane such that AB =12v (of course, either A, or
B, may be chosen arbitrarily). �Remark : Since a translation is a product of two central symmetries, then
the translation is an isometry.
87
Figure 4.7:
Figure 4.8:
Proposition 4.3.2. A product of two axial symmetries, whose axes are pa-rallel, is a translation. Conversely, any translation can be decomposed into aproduct of two axial symmetries, having parallel axes.
Proof:Let d and d′ be two parallel lines. Let M be an arbitrary point on the
plane, M ′ = sd(M) and M ′′ = sd′(M ′) = (sd′ ◦ sd)(M) (see Figure 4.8).If {A} = MM ′ ∩ d and {B} = M ′M ′′ ∩ d′, then MM ′′ = 2AB. One hassd′ ◦ sd = tAB, where the vector AB is orthogonal on d and its length is a halfof the distance between the lines d and d′.
Conversely, given a translation tv, one can choose an arbitrary line d,orthogonal on the direction of v. Let A be an arbitrary point on d and there
exists a unique point B, such that AB =12v. Let d′ be the line passing through
B and parallel to d. Then tv = sd′ ◦ sd. �
Proposition 4.3.3. Given two points A and A′ in E2, there exists a uniquetranslation tv, such that tv(A) = A′. In particular, if a translation tv has afixed point, then tv = 1E2.
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Proof: tAA′ is the unique translation having the required property. Iftv(P ) = P , then v = PP = 0, and tv = t0 = 1E2 . �
Denote byT (E2) = {tv : E2 → E2 : tv translation },
the set of all the translations of the plane.
Theorem 4.3.4. (T (E2), ◦) is a commutative subgroup of (Iso(E2), ◦), iso-morphic to (V2,+).
Proof: It is already settled that T (E2) ⊂ Iso(E2). If tv and tw are twotranslations, it easy to see that
tv + tw = tv+w and t−1v = t−v, (4.1)
so that the required isomorphism is natural: f(tv) = v. �
Theorem 4.3.5. Let a, b and c be three parallel lines on E2. Then sa ◦ sb ◦ sc
is an axial symmetry, whose axis is parallel to the three given lines.
Proof: The product sb ◦ sc is a translation (see Proposition 4.3.2). Thistranslation is the product of two axial symmetries, and one may choose oneof them to be sa and the other to be sd, where d is parallel to a. Then
sa ◦ sb ◦ sc = sa ◦ (sb ◦ sc) = sa ◦ (sa ◦ sd) = (sa ◦ sa) ◦ sd = sd. �
4.3.1 Exercises
1) Let C(O,R) be a circle and A,B ∈ C(O,R) two fixed points. Find thegeometric locus of the orthocenter of the triangle ∆ABC, when C is amobile point on C(O,R).
2) Let d be a line, A and B be two points situated on the same half-planewith respect to d and a ≥ 0 a given number. Determine two points Mand N on the line d, such that:
a) |MN | = a;
b) |AM |+ |MN |+ |NB| is minimum.
3) Construct a trapezium, knowing the lengths of its sides.
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4) Let C1(O1, r1) and C2(O2, r2) be two circles and d be an arbitrary line.Construct a line d′, parallel to d, such that the chords determined by d′
on the two circles are of the same length.
5) Let C1(O1, r1) and C2(O2, r2) be two circles and [MN ] a segment. Con-struct a segment [AB], parallel to [MN ] and of the same length, suchthat A ∈ C1(O1, r1) and B ∈ C2(O2, r2).
6) Prove that the group (T (E2), ◦) is isomorphic to the additive group ofthe complex numbers.
4.4 Rotations
An angle AOB is said to be oriented if the pair of half-lines {[OA, [OB}is ordered. The angle AOB is positively oriented if [OA gets over [OB coun-terclockwisely. Otherwise, AOB is negatively oriented. If the measure of thenonoriented angle AOB is θ, then the measure of the oriented angle AOB iseither θ, or −θ, depending on the orientation of AOB.
Figure 4.9:
Let O ∈ E2 be a point and θ ∈ [−2π, 2π] be a number. The maprO,θ : E2 → E2, given by
rO,θ(M) = M ′, where
{|OM | = |OM ′|m(MOM ′) = θ
,
is called rotation of center O and oriented angle θ.The point O is the center of the rotation and the number θ is the angle of
rotation.
Remarks:
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Figure 4.10:
• If θ ∈ {−2π, 0, 2π}, then the rotation of angle θ is the identity of theplane, rO,−2π = 1E2 , rO,0 = 1E2 , rO,2π = 1E2 ;
• If θ ∈ {−π, π}, then the rotation of angle θ is the central symmetry,rO,−π = sO, rO,π = sO;
• The rotation rO,θ is an involution if and only if θ ∈ {−2π,−π, 0, π, 2π};
• The center O of the rotation rO,θ (of an angle θ 6= −2π, 0, 2π) is itsunique fixed point and there are no invariant lines with respect to rO,θ.
Theorem 4.4.1. Any rotation is an isometry of E2.
Proof: Let rO,θ be a rotation of center O, let A and B be two arbitrarypoints in E2 and A′ = rO,θ(A), B′ = rO,θ(B). Using the definition of the
Figure 4.11:
rotation, one has that [OA] ≡ [OA′], [OB] ≡ [OB′] and AOB ≡ A′OB′.Therefore, ∆AOB ≡ ∆A′OB′ so that |AB| = |A′B′|. �
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Theorem 4.4.2. Any isometry with a unique fixed point O is a rotation ofcenter O.
Proof: Let f ∈ Iso(E2) and f(O) = O. Let M ∈ E2 \ {O} be an arbitrarypoint and M ′ = f(M), M ′ 6= M . There exist exactly two isometries suchthat f(O) = O and f(M) = M ′ (see Theorem 4.2.4) and, since the rotation ofcenter O and angle MOM ′ and the axial symmetry with respect to the internalbisector of the angle MOM ′ have these properties, and since the latter has anentire line of fixed points, then f must be a rotation of center O. �
Remarks:
• Sometimes, the Theorem 4.4.2 is given as definition of rotation: therotation is the isometry having a unique fixed point, called center, orthe identity 1E2 ;
• Let us denote by R(O) the set of all the rotations of center O and angleθ ∈ [−2π, 2π]. Then (R(O), ◦) is a commutative subgroup of (Iso(E2), ◦).Indeed, rO,θ1 ◦ rO,θ2 = rO,θ1+θ2( mod 2π) and r−1
O,θ = rO,−θ.
Proposition 4.4.3. A product of two axial symmetries, whose axis are con-current, is a rotation. Conversely, any rotation can be expressed as a productof two axial symmetries, whose axis are concurrent at the center of rotation.
Proof: Let sa and sb be two axial symmetries, with a∩ b = {O}. We shallprove that sb ◦ sa is a rotation of center O, by verifying that O is the uniquefixed point of sb ◦ sa. First,
(sb ◦ sa)(O) = sb(sa(O)) = sb(O) = O,
hence O is a fixed point.Let M be another fixed point of sb ◦ sa. Since (sb ◦ sa)(M) = M , then
(s−1b ◦ sb ◦ sa)(M) = s−1
b (M) and, because the symmetry sb is involutive, itfollows that sa(M) = sb(M) = N . If the pointsM andN are distinct, M 6= N ,then the lines a and b are both perpendicular on the midpoint of the segment[MN ], contradiction to the fact that a ∩ b = {O}. Hence M = N . ThenM = sa(M) and M = sb(M), meaning that M ∈ a and M ∈ b. It follows,necessarily, that M = O, and O is the unique fixed point of sb ◦ sa.
Conversely, denote r = rO,θ and let A 6= O be an arbitrary point andA′ = r(A). Let b be the bisector of the angle AOA′. The map sb ◦ r is anisometry and, moreover,
(sb ◦ r)(O) = sb(r(O)) = sb(O) = O
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and(sb ◦ r)(A) = sb(r(A)) = sb(A′) = A,
so that sb ◦ r has at least two fixed points, A and O. Therefore, any pointof OA is fixed and sb ◦ r is either the identity, or the axial symmetry withrespect to a = OA (see Theorem 4.2.3). If sb ◦ r = 1E2 , then sb = r, whichis impossible, since r has a unique fixed point and sb has an infinity of fixedpoints. Then, sb ◦ r = sa, and r = sb ◦ sa.
Because the point A was chosen arbitrarily, one of the axis a and b can bechosen arbitrarily. �
Remark : It is easy to see, in the previous theorem, that the angle ofrotation is twice the (oriented) angle aOb. Then, the product of two axialsymmetries, having orthogonal axes a⊥b, is a central symmetry of center O,with {O} = a ∩ b.
Theorem 4.4.4. Let a, b and c be three lines on E2, such that a∩b∩c = {O}.Then sa ◦ sb ◦ sc is an axial symmetry, whose axis d passes through O.
Proof: Use the Proposition 4.4.3. The product sb◦sc is a rotation of centerO. This rotation can be decomposed into a product of two axial symmetries,the axes passing through O, and one of them can be chosen arbitrarily, so thatit can be a. Then sb ◦ sc = sa ◦ sd and
sa ◦ sb ◦ sc = sa ◦ sa ◦ sd = sd. �
4.4.1 Exercises
1) Let ∆ABC be an equilateral triangle and M a point in the same plane.
a) Prove that, if M /∈ C(A,B,C), then there exists a triangle havingthe sides of lengths |MA|, |MB|, |MC|.
b) Prove that, if M ∈ C(A,B,C), then the length of one of the seg-ments [MA], [MB], [MC] is the sum of the lengths of the twoothers.
2) Let ∆ABC and ∆DBC be two equilateral triangles, symmetrical withrespect to BC, and M a point on the circle of center D and radius |DB|.If a, b, c are the lengths of the segments [BC], [CA], respectively [AB],prove that a2 = b2 + c2.
3) Let a, b and c be three parallel lines. Construct an equilateral triangle∆ABC, such that A ∈ a, B ∈ b, C ∈ c.
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4) Let C1, C2, C3 be three circles, of radius r1 < r2 < r3, respectively.Determine an equilateral triangle ∆ABC, such that A ∈ a, B ∈ b,C ∈ c.
5) Determine a point M in the plane of the triangle ∆ABC, such that|MA|+ |MB|+ |MC| is minimal.
6) Let d be a line and A /∈ d. Find the geometric locus of the vertex C ofan equilateral triangle ∆ABC, when B ∈ d.
7) Prove that the product of two rotations is commutative if and only ifthe centers of the rotations coincide.
4.5 Analytic Form of Isometries
In this paragraph, we determine the equations of the plane isometries. LetxOy be a Cartesian system of coordinates, associated to E2.
Proposition 4.5.1. Let P (x0, y0) be the center of the central symmetry sP .The equations of sP are {
x′ = 2x0 − xy′ = 2y0 − y
. (4.2)
Proof: Let M(x, y) be an arbitrary point on E2 and M ′ = sP (M) itssymmetrical with respect to P , M ′ = (x′, y′).
Figure 4.12:
Since P is the midpoint of the segment [MM ′], then x0 =x+ x′
2and
y0 =y + y′
2, and the equations (4.2) hold. �
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Remark : If the center of symmetry is the origin O(0, 0) of the system ofcoordinates, then the equations (4.2) become{
x′ = −xy′ = −y . (4.3)
Proposition 4.5.2. Let d : ax+ by+ c = 0, a2 + b2 > 0, be a line in E2. Theequations of the axial symmetry sd are:
x′ =b2 − a2
a2 + b2x−
2aba2 + b2
y −2ac
a2 + b2
y′ = −2ab
a2 + b2x−
b2 − a2
a2 + b2y −
2bca2 + b2
. (4.4)
Proof: One may suppose that b 6= 0. Let M(x, y) be an arbitrary pointand M ′ = sd(M), M ′(x′, y′).
Figure 4.13:
The points M and M ′ are symmetrical with respect to d if and only if theline passing through M and M ′ is orthogonal on d and the midpoint P of thesegment [MM ′] belongs to d.
The equation of the line determined by M and M ′ isX − x
x′ − x=Y − y
y′ − y. The
orthogonality condition gives a(y′ − y) = b(x′ − x).The midpoint of [MM ′] is a point of d if and only if
a
(x+ x′
2
)+ b
(y + y′
2
)+ c = 0.
Then, the coordinates (x′, y′) of M ′ are the solution of the system of equa-tion {
ax′ + by′ = −(ax+ by + 2c)bx′ − ay′ = bx− ay
and one obtains (4.4). �
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Remarks:
• If the line d passes through the origin O, then c = 0 and (4.4) becomesx′ =
b2 − a2
a2 + b2x−
2aba2 + b2
y
y′ = −2ab
a2 + b2x−
b2 − a2
a2 + b2y
. (4.5)
• If the line d is parallel to Ox, then a = 0 and (4.4) turns into x′ = x
y′ = −y −2cb
. (4.6)
• If the line d is parallel to Oy, then b = 0 and (4.4) converts into x′ = −x−2ca
y′ = y. (4.7)
Proposition 4.5.3. Let v(x0, y0) be a vector. The equations of the translationtv of vector v are {
x′ = x+ x0
y′ = y + y0. (4.8)
Proof:
Figure 4.14:
If M(x, y) is an arbitrary point and M ′ = tv(M), M ′(x′, y′), thenMM ′ = v. By identifying the components of the vectors, one obtains theequations (4.8). �
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Proposition 4.5.4. Let f ∈ Iso(E2) be an isometry, having the origin O(0, 0)as fixed point. The equations of f are{
x′ = ax− εbyy′ = bx+ εay
, (4.9)
where a2 + b2 = 1 and ε = ±1.Conversely, any system of equation of the form (4.9) represents an iso-
metry of E2, having O as fixed point.
Proof: Take the point A(1, 0) and let A′ = f(A), with A′(a, b). Let M(x, y)be an arbitrary point and M ′ = f(M), M ′(x′, y′). Since 1 = |OA| = |OA′|,then a2 + b2 = 1.On the other hand,{
|OM | = |OM ′||AM | = |A′M ′| ⇐⇒
{x2 + y2 = x′2 + y′2
(x− 1)2 + y2 = (x′ − a)2 + (y′ − b)2⇐⇒
⇐⇒{
x = ax′ + by′
x2 + y2 = x′2 + y′2
Replacing x in the second equation, one gets
(ax′ + by′)2 + y2 = x′2 + y′2,
or
y2 = (1−a2)x′2− 2abx′y′+(1− b2)y′2 = b2x′2− 2abx′y′+a2y′2 = (bx′−ay′)2.
One obtains the system of equations{x = ax′ + by′
εy = −bx′ + ay′, ε = ±1,
which gives (4.9).Conversely, it is clear that O(0, 0) is a fixed point, since, for x = y = 0,
one obtains x′ = y′ = 0.Let M1(x1, y1) and M2(x2, y2) be two arbitrary points and M ′
1(x′1, y
′1),
M ′2(x
′2, y
′2) be their images, whose coordinates are given through (4.9). Then
|M ′1M
′2|2 = (x′2 − x′1)
2 + (y′2 − y′1)2 =
= [a(x2 − x1)− εb(y2 − y1)]2 + [b(x2 − x1) + εa(y2 − y1)]2 =
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= (a2 + b2)[(x2 − x1)2 + (y2 − y1)2] = |M1M2|2,
therefore |M ′1M
′2| = |M1M2| and the equations (4.9) define an isometry. �
Let f ∈ Iso(E2) be an arbitrary isometry of E2. Suppose that f(O) = O′.The product g = t−1
OO′ ◦ g is an isometry and g(O) = O, therefore O is a fixedpoint for g. Using Propositions 4.5.3 and 4.5.4, one obtains:
Theorem 4.5.5. The equations of an arbitrary isometry of E2 are{x′ = ax− εby + x0
y′ = bx+ εay + y0, (4.10)
where a2 + b2 = 1 and ε = ±1.
Remark : The equations (4.10) can be written in the form(x′
y′
)=(a −εbb εa
)(xy
)+(x0
y0
), (4.11)
where a2 + b2 = 1 and ε = ±1.The isometries can be characterized through their fixed points. A point
M(x, y) is fixed if and only if its coordinates satisfy the system{(a− 1)x− εby = −x0
bx+ (εa− 1)y = −y0. (4.12)
The discriminant of (4.12) is ∆ =∣∣∣∣a− 1 −εbb εa− 1
∣∣∣∣ = (1− a)(ε− 1).
• If ε = −1, then (4.12) has an infinity of solutions, and f is the productbetween an axial symmetry and a translation;
• If ε = 1, a 6= 1, then (4.12) has a unique solution, and f is the productbetween a rotation of angle θ = arccos a, centered at the fixed point off , and a translation. The equations of f become(
x′
y′
)=(
cos θ − sin θsin θ cos θ
)(xy
)+(x0
y0
); (4.13)
• If ε = 1, a = 1, then b = 0 and f is a translation.
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4.5.1 Exercises
1) Let A(−1, 2) and B(2,−1) be two points and d : 2x+ 3y − 6 = 0 a lineon E2, let v(2, 3) be a vector, C = tv(A), D = tv(B) and b = tv(a).
a) Find the coordinates of C, D and the equation of the line b;
b) Find the coordinates of the intersection between CD and b.
2) Let A(2, 1) and B(6, 4) be two points and a : x = 3, b : x = 4 be twolines on E2.
a) Find two points M ∈ a and N ∈ b, such that |AM | = |BN | andthe line MN is parallel to the first bisector x− y = 0;
b) Find the points P ∈ a, Q ∈ b, such that |AP | + |PQ| + |QB| isminimal.
3) Let a be a fixed line and A ∈ a be fixed. A variable circle of given radiusr passes through A and intersects the line a at M . Let P be a point onthe circle, such that [AP ] ≡ [MP ].
a) Find the geometric locus of P ;
b) Find the geometric locus of the orthocenter of the triangle ∆AMP .
4) Find the symmetrical of the point P (4, 2) with respect to the lined : x+ 2y − 4 = 0.
5) Find the coordinates of the symmetrical of M(3,−2) with respect toP (2, 1).
6) Find the coordinates of a point M ∈ Ox, such that the sum of itsdistances to A(1, 2) and B(3, 4) is minimum.
7) The point A(−4, 5) is the vertex of a square having a diagonal containedinto the line d : 7x − y + 8 = 0. Find the coordinates of the vertices ofthe square.
8) Find the equations of the symmetricals of the line d : 6x + 5y − 15 = 0with respect to Ox, Oy and O.
9) Find the equation of the symmetrical of the line d : −x + 2y − 1 = 0with respect to a : x− y = 0.
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100
Chapter 5
Conics
5.1 The Circle
5.1.1 Definition
A circle is a closed plane curve, defined as the geometric locus of the pointsat a given distance R from a point I. The point I is the center of the circleand the number R is the radius of the circle. We shall denote the circle ofcenter I and radius R by C(I,R).
In order to determine the equation of the circle, suppose that xOy is anassociated Cartesian system of coordinates in E2, and I(a, b). An arbitrarypoint M(x, y) belongs to C(I,R) if and only if |MI| = R.
Figure 5.1:
Hence,√
(x− a)2 + (y − b)2 = R, or
(x− a)2 + (y − b)2 = R2. (5.1)
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The equation (5.1) represents the equation of the circle centered at I(a, b) andof radius R.
Remark : In a Cartesian system of coordinates, the equation
x2 + y2 − 2ax− 2by + c = 0 (5.2)
represents either a circle, or a point, or the empty set. Indeed, (5.2) can bewritten in the equivalent form
(x− a)2 + (y − b)2 = a2 + b2 − c.
• If a2 + b2− c > 0, then (5.2) is the equation of the circle of center I(a, b)and radius R =
√a2 + b2 − c;
• If a2 + b2 − c = 0, then (5.2) represents the point I(a, b) (the circle isdegenerated to its center);
• If a2 + b2 − c < 0, then (5.2) is the empty set (or an imaginary circle).
The equation (5.2), x2 + y2 − 2ax − 2by + c = 0, with a2 + b2 − c > 0, issaid to be the general equation of the circle.
5.1.2 The Circle Determined by Three Points
Given three noncollinear pointsM1(x1, y1), M2(x2, y2) andM3(x3, y3), thereexists a unique circle passing through them. Suppose that the circle deter-mined by M1(x1, y1), M2(x2, y2) and M3(x3, y3) has the general equation
x2 + y2 − 2ax− 2by + c = 0,
with a2 + b2 − c = 0. Since the three points are on the circle, one obtains thesystem of equations (with variables a, b and c)
x2 + y2 − 2ax− 2by + c = 0x2
1 + y21 − 2ax1 − 2by1 + c = 0
x22 + y2
2 − 2ax2 − 2by2 + c = 0x2
3 + y23 − 2ax3 − 2by3 + c = 0
,
which has to be compatible, so that∣∣∣∣∣∣∣∣x2 + y2 x y 1x2
1 + y21 x1 y1 1
x22 + y2
2 x2 y2 1x2
3 + y23 x3 y3 1
∣∣∣∣∣∣∣∣ = 0. (5.3)
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The equation (5.3) is the equation of the circle determined by three points.It follows immediately that four points Mi(xi, yi), i = 1, 4, belong to a
circle if and only if ∣∣∣∣∣∣∣∣x2
1 + y21 x1 y1 1
x22 + y2
2 x2 y2 1x2
3 + y23 x3 y3 1
x24 + y2
4 x4 y4 1
∣∣∣∣∣∣∣∣ = 0. (5.4)
5.1.3 Intersection of a Circle and a Line
Let C be a circle and d be a line on E2. One may choose a system ofcoordinates having the center at the center of the circle, so that the equationof C is x2 + y2 −R2 = 0. Let d : y = mx+ n.
The intersection between C and d is given by the solutions of the systemof equations {
x2 + y2 −R2 = 0y = mx+ n
.
By substituting y in the equation of the circle, one obtains
(1 +m2)x2 + 2mnx+ n2 −R2 = 0.
The discriminant of this second degree equation is
∆ = 4(R2 +m2R2 − n2).
• If R2 +m2R2 − n2 < 0, then there are no intersection points between Cand d. The line is exterior to the circle;
• If R2 +m2R2 − n2 = 0, then there is a double point (a tangency point)between C and d. The line is tangent to the circle. The coordinates of
the tangency point are
(−
mn
1 +m2,
n
1 +m2
);
• If R2 +m2R2 − n2 > 0, then there are two intersection points betweenC and d. The line is secant to the circle. If x1 and x2 are the roots ofthe above equation, then the intersection points between C and d areP1(x1,mx1 + n) and P2(x2,mx2 + n).
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5.1.4 The Tangent to a Circle
The Tangent Having a Given Direction
Let C be the circle of equation x2 + y2 − R2 = 0 and m ∈ R a given realnumber. There are two lines, having the angular coefficient m, and which aretangent to C. We saw, in the previous paragraph, that a line d : y = mx+ nis tangent to C if and only if R2 +m2R2 − n2 − 0. Then, the equations of thetwo tangent lines of direction m are
y = mx±R√
1 +m2. (5.5)
The Tangent to a Circle at a Point of the Circle
Let C : x2 + y2 − r2 = 0 be a circle and P0(x0, y0) be a point on C. Thetangent at P0 to C is a line from the bundle of lines y−y0 = m(x−x0), m ∈ R,having the vertex P . On the other hand, the tangent has to be of the form(5.5): y = mx±R
√1 +m2. Then, the angular coefficient m must verify{
y − y0 = m(x− x0)y = mx±R
√1 +m2 ,
hence(y0 −mx0)2 = R2(1 +m2).
But x20 + y2
0 = R2 (since P0 ∈ C) and one obtains (mx0 − y0)2 = 0. Therefore
m = −x0
y0(one may suppose that y0 6= 0; otherwise, one gets the tangent at
the point (R, 0), which is of equation x = R). Replacing m in the equation ofthe bundle, one obtains
y − y0 = −x0
y0,
orx0x+ y0y − (x2
0 + y20) = 0.
Again, x20 + y2
0 = R2, and the equation of the tangent line to C at the pointP0 ∈ C is
x0x+ y0y −R2 = 0. (5.6)
Remark : The equation of the line OP0 is y =y0
x0x. Then, the product of
the angular coefficients of OP0 and of the tangent at P0 is −1, meaning thatthe tangent at a point to a circle is orthogonal on the radius which correspondsto the point.
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5.1.5 Intersection of Two Circles
Given two circles,
C1 : x2 + y2 − 2a1x− 2b1y + c1 = 0
andC2 : x2 + y2 − 2a2x− 2b2y + c2 = 0,
the system of equations{x2 + y2 − 2a1x− 2b1y + c1 = 0x2 + y2 − 2a2x− 2b2y + c2 = 0
gives informations about the intersection of the two circles. The previoussystem is equivalent to{
x2 + y2 − 2a1x− 2b1y + c1 = 02(a2 − a1)x+ 2(b2 − b1)y − (c2 − c1) = 0
which will give rise to a second degree equation, having the discriminant ∆.
• If ∆ > 0, then C1 and C2 are secant (they have two intersection points);
• If ∆ = 0, then C1 and C2 are tangent (they have one tangency point);
• If ∆ < 0, then C1 and C2 have no intersection points.
5.1.6 Exercises
1) Find the equation of the circle of diameter [AB], where A(1, 2) andB(−3,−1).
2) Find the equation of the circle of center I(2,−3) and radius R = 7.
3) Find the equation of the circle of center I(−1, 2) and which passesthrough A(2, 6).
4) Find the equation of the circle centered at the origin and tangent tod : 3x− 4y + 20 = 0.
5) Find the equation of the circle passing through A(3, 1) and B(−1, 3) andhaving the center on the line d : 3x− y − 2 = 0.
6) Find the equation of the circle determined by A(1, 1), B(1,−1) andC(2, 0).
105
7) Find the equation of the circle tangent to both d1 : 2x+ y − 5 = 0 andd2 : 2x+ y + 15 = 0, if the tangency point with d1 is M(3, 1).
8) Determine the position of the point A(1,−2) relative to the circleC : x2 + y2 − 8x− 4y − 5 = 0.
9) Find the intersection between the line d : 7x− y + 12 = 0 and the circleC : (x− 2)2 + (y − 1)2 − 25 = 0.
10) Determine the position of the line d : 2x−y−3 = 0 relative to the circleC : x2 + y2 − 3x+ 2y − 3 = 0.
11) Find the equation of the tangent to C : x2 + y2 − 5 = 0 at the pointA(−1, 2).
12) Find the equations of the tangent lines to C : x2 + y2 +10x− 2y+6 = 0,parallel to d : 2x+ y − 7 = 0.
13) Find the equations of the tangent lines to C : x2 + y2 − 2x + 4y = 0,orthogonal on d : x− 2y + 9 = 0.
14) Let Cλ : x2 +y2 +λx+(2λ+3)y = 0, λ ∈ R, be a family of circles. Provethat the circles from the family have two fixed points.
15) Find the geometric locus of the points in the plane for which the sumof the squares of the distances to the sides of an equilateral triangle isconstant.
16) Let P and Q be two fixed points and d a line, orthogonal on PQ. Twovariable orthogonal lines, passing through P , cut d at A, respectively B.Find the geometric locus of the orthogonal projection of the point A onthe line BQ.
17) Two circles of centers O, respectively O′, intersect each other at A and B.A variable line passing through A cuts the two circles at C, respectivelyC ′. Find the geometric locus of the intersection point between the linesOC and O′C ′.
5.2 The Ellipse
5.2.1 Definition
An ellipse is a plane curve, defined to be the geometric locus of the pointsin the plane, whose distances to two fixed points have a constant sum.
106
The two fixed points are called the foci of the ellipse and the distancebetween the foci is the focal distance.
Let F and F ′ be the two foci of an ellipse and let |FF ′| = 2c be the focaldistance. Suppose that the constant in the definition of the ellipse is 2a. If Mis an arbitrary point of the ellipse, it must verify the condition
|MF |+ |MF ′| = 2a.
One may chose a Cartesian system of coordinates centered at the midpoint ofthe segment [F ′F ], so that F (c, 0) and F ′(−c, 0).
Figure 5.2:
Remark : In ∆MFF ′ holds |MF |+ |MF ′| > |FF ′|, hence 2a > 2c. Then,the constants a and c must verify a > c.
Let us determine the equation of an ellipse. Starting with the definition,|MF |+ |MF ′| = 2a, or√
(x− c)2 + y2 +√
(x+ c)2 + y2 = 2a.
This is equivalent to√(x− c)2 + y2 = 2a−
√(x+ c)2 + y2
and(x− c)2 + y2 = 4a2 − 4a
√(x+ c)2 + y2 + (x+ c)2 + y2.
One obtainsa√
(x+ c)2 + y2 = cx+ a2,
which gives
a2(x2 + 2xc+ c2) + a2y2 = c2x2 + 2a2cx+ a2,
107
or(a2 − c2)x2 + a2y2 − a2(a2 − c2) = 0.
Denoting a2 − c2 = b2 (possible, since a > c), one has
b2x2 + a2y2 − a2b2 = 0.
Dividing by a2b2, one obtains the equation of the ellipse
x2
a2+y2
b2− 1 = 0. (5.7)
Remark : The equation (5.7) is equivalent with
y = ±b
a
√a2 − x2; x = ±
a
b
√b2 − y2,
which means that the ellipse is symmetrical with respect to both Ox andOy. In fact, the line FF ′, determined by the foci of the ellipse, and theperpendicular line on the midpoint of the segment [FF ′] are axes of symmetryfor the ellipse. Their intersection point, which is the midpoint of [FF ′], is thecenter of symmetry of the ellipse, or, simply, its center .
In order to sketch the graph of the ellipse, remark that is it enough torepresent the function
f : [−a, a] → R, f(x) =b
a
√a2 − x2,
and to complete the ellipse by symmetry with respect to Ox.One has
f ′(x) = −b
a
x√a2 − x2
, f ′′(x) = −ab
(a2 − x2)√a2 − x2
.
x −a 0 a
f ′(x) | + + + 0 − − − |f(x) 0 ↗ b ↘ 0f ′′(x) | − − − − − − − |
The graph of the ellipse is presented in Figure 5.3.
108
Figure 5.3:
Remarks:
• In particular, if a = b in (5.7), one obtains the equation x2 + y2−a2 = 0of the circle centered at the origin and of radius a. This happens whenc = 0, i.e. when the foci coincide, so that the circle may be seen as anellipse whose foci are identical.
• All the considerations can be done in a similar way, by taking the fociof the ellipse on Oy. One obtains a similar equation for such an ellipse.
The number e =c
ais called the eccentricity of the ellipse. Since a > c,
then 0 < e < 1, hence any ellipse has the eccentricity smaller that 1. On the
other hand, e2 =c2
a2= 1−
(b
a
)2
, so that e gives informations about the shape
of the ellipse. When e is closer and closer to 0, then the ellipse is ”closer andcloser” to a circle; and when e is closer to 1, then the ellipse is flattened toOx.
5.2.2 Intersection of a Line and an Ellipse
Given an ellipse E :x2
a2+y2
b2− 1 = 0 and a line d : y = mx + n, their
intersection is given by the solutions of the system of equations x2
a2+y2
b2− 1 = 0
y = mx+ n,
109
or, by replacing y in the equation of the ellipse,
(a2m2 + b2)x2 + 2a2mnx+ a2(n2 − b2) = 0.
The discriminant ∆ of the last equation is given by
∆ = 4[a4m2n2 − a2(a2m2 + b2)(n2 − b2)].
• If ∆ < 0, then d does not intersect E . The line is exterior to the ellipse;
• If ∆ = 0, then the line is tangent to the ellipse. There is a tangencypoint between d and E ;
• If ∆ > 0, then there are two intersection points between d and E . Theline is secant to the ellipse.
5.2.3 The Tangent to an Ellipse
The Tangent of a Given Direction
If E :x2
a2+y2
b2− 1 = 0 is an ellipse and m ∈ R a given real number,
there exist exactly two lines, having the angular coefficient m and tangentto E . Since a line d : y = mx + n is tangent to the ellipse if and only ifa4m2n2− a2(a2m2 + b2)(n2− b2) = 0, then n = ±
√a2m2 + b2. The equations
of the tangent lines of direction m are
y = mx±√a2m2 + b2. (5.8)
The Tangent to an Ellipse at a Point of the Ellipse
Let E :x2
a2+y2
b2−1 = 0 be an ellipse and P0(x0, y0) be a point of E . Suppose
that y0 > 0, so that P0 is situated on the graph of the function f : [−a, a] → R,
f(x) =b
a
√a2 − b2. The angular coefficient of the tangent at P0 to E is
f ′(x0) = −b
a
x0√a2 − b2
= −b2x0
a2y0.
If y0 < 0, a similar argument shows that the angular coefficient of the
tangent at P0 is still −b2x0
a2y0.
110
The equation of the tangent to E at P0 is
y − y0 = −b2x0
a2y0(x− x0),
equivalent tob2x0(x− x0) + a2y0(y − y0) = 0,
orx0x
a2+y0y
b2−
(x2
0
a2+y20
b2
)= 0.
Since P0 belongs to the ellipse, thenx2
0
a2+y20
b2= 1, and the equation of the
tangent to the ellipse at the point P0 is
x0x
a2+y0y
b2− 1 = 0. (5.9)
5.2.4 Exercises
1) Determine the coordinates of the foci of the ellipse 9x2 +25y2−225 = 0.
2) Sketch the graph of y = −34
√16− x2.
3) Find the intersection points between the line x + 2y − 7 = 0 and theellipse x2 + 3y2 − 25 = 0.
4) Find the position of the line 2x + y − 10 = 0 relative to the ellipsex2
9+y2
4− 1 = 0.
5) Find the equation of the tangent to the ellipse E : x2 + 4y2 − 20 = 0,orthogonal on the line d : 2x− 2y − 13 = 0.
6) Find the equations of the tangent lines atx2
25+y2
16− 1 = 0, passing
through P0(10,−8).
7) Find the geometric locus of the orthogonal projections of a focus of anellipse on the tangent lines to the ellipse.
8) Let d1 and d2 be two variable orthogonal lines, passing through the point
A(a, 0) of the ellipsex2
a2+y2
b2= 1, and let P1 and P2 be the intersection
points of these two lines and the ellipse. Prove that the line P1P2 passesthrough a fixed point.
111
5.3 The Hyperbola
5.3.1 Definition
A hyperbola is a plane curve, defined as the geometric locus of the points inthe plane, whose distances to two fixed points have a constant difference.
The two fixed points are called the foci of the hyperbola, and the distancebetween the foci is the focal distance.
Denote by F and F ′ the foci of the hyperbola and let |FF ′| = 2c be thefocal distance. Suppose that the constant in the definition is 2a. If M(x, y) isan arbitrary point of the hyperbola, then
||MF | − |MF ′|| = 2a.
Choose a Cartesian system of coordinates, having the center at the mid-point of the segment [FF ′] and such that F (c, 0), F ′(−c, 0).
Figure 5.4:
Remark : In the triangle ∆MFF ′, ||MF | − |MF ′|| < |FF ′|, so that a < c.The metric relation |MF | − |MF ′| = ±2a becomes√
(x− c)2 + y2 −√
(x+ c)2 + y2 = ±2a,
or, √(x− c)2 + y2 = ±2a+
√(x+ c)2 + y2.
This is
x2 − 2cx+ c2 + y2 = 4a2 ± 4a√
(x+ c)2 + y2 + x2 + 2cx+ c2 + y2 ⇐⇒
⇐⇒ cx+ a2 = ±a√
(x+ c)2 + y2 ⇐⇒
⇐⇒ c2x2 + 2a2cx+ a4 = a2x2 + 2a2cx+ a2c2 + a2y2 ⇐⇒
112
⇐⇒ (c2 − a2)x2 − a2y2 − a2(c2 − a2) = 0.
Denote c2 − a2 = b2 (possible, since c > a) and one obtains the equation ofthe hyperbola
x2
a2−y2
b2− 1 = 0. (5.10)
Remark : The equation (5.10) is equivalent to
y = ±b
a
√x2 − a2; x = ±
a
b
√y2 + b2.
Then, the coordinate axes are axes of symmetry for the hyperbola. Theirintersection point is the center of the hyperbola.
To sketch the graph of the hyperbola, is it enough to represent the function
f : (−∞,−a] ∪ [a,∞) → R, f(x) =b
a
√x2 − a2,
by taking into account that the hyperbola is symmetrical with respect to Ox.
Since limx→∞
f(x)x
=b
aand lim
x→−∞
f(x)x
= −b
a, it follows that y =
b
ax and
y = −b
ax are asymptotes of f .
One has, also,
f ′(x) =b
a
x√x2 − a2
, f ′′(x) = −ab
(x2 − a2)√x2 − a2
.
x −∞ −a a ∞f ′(x) − − − − | � � � | + + + +f(x) ∞ ↘ 0| � � � |0 ↗ ∞f ′′(x) − − − − | � � � | − − − −
The graph of the hyperbola is presented in Figure 5.5.
Remarks:
• If a = b, the equation of the hyperbola becomes x2 − y2 = a2. In thiscase, the asymptotes are the bisectors of the system of coordinates andone deals with an equilateral hyperbola.
• As in the case of an ellipse, one can consider the hyperbola having thefoci on Oy.
113
Figure 5.5:
The number e =c
ais called the eccentricity of the hyperbola. Since c > a,
then the eccentricity is always greater than 1. Moreover,
e2 =c2
a2= 1 +
(b
a
)2
,
hence e gives informations about the shape of the hyperbola. For e closer to1, the hyperbola has the branches closer to Ox.
5.3.2 Intersection of a Hyperbola and a Line
Let H :x2
a2−y2
b2− 1 = 0 be a hyperbola and d : y = mx+ n be a line in E2.
Their intersection is given by the system of equations x2
a2−y2
b2− 1 = 0
y = mx+ n.
By substituting y in the first equation, one obtains
(a2m2 − b2)x2 + 2a2mnx+ a2(n2 + b2) = 0. (5.11)
• If a2m2 − b2 = 0, (or m = ±b
a), then the equation (5.11) becomes
±2bnx+ a(n2 + b2) = 0.
114
– If n = 0, there are no solutions (this means, geometrically, that thetwo asymptotes do not intersect the hyperbola);
– If n 6= 0, there exists a unique solution (geometrically, a line d,which is parallel to one of the asymptotes, intersects the hyperbolaat exactly one point);
• If a2m2 = b2 6= 0, then the discriminant of the equation (5.11) is
∆ = 4[a4m2n2 − a2(a2m2 − b2)(n2 + b2)].
– If ∆ < 0, then the line does not intersect the hyperbola;
– If ∆ = 0, then the line is tangent to the hyperbola (they have adouble intersection point);
– If ∆ > 0, then the line and the hyperbola have two intersectionpoints.
5.3.3 The Tangent to a Hyperbola
The Tangent of a Given Direction
The line d : y = mx + n is tangent to the hyperbola H :x2
a2−y2
b2− 1 = 0
if the discriminant ∆ of the equation (5.11) is zero, which is equivalent toa2m2 − n2 − b2 = 0.
• If a2m2−b2 ≥ 0, i.e. m ∈
(−∞,−
b
a
]∪
[b
a,∞
), then n = ±
√a2m2 − b2.
The equations of the tangent lines to H, having the angular coefficientm are
y = mx±√a2m2 − b2. (5.12)
• If a2m2 − b2 < 0, there are no tangent lines to H, of angular coefficientm.
The Tangent at a Point of the Hyperbola
One can prove, as in the case of the ellipse that, if H :x2
a2−y2
b2− 1 = 0 is a
hyperbola, and P0(x0, y0) is a point of H, then the equation of the tangent toH at P0 is
x0x
a2−y0y
b2− 1 = 0. (5.13)
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5.3.4 Exercises
1) Find the foci of the hyperbolax2
9−y2
4− 1 = 0.
2) Find the area of the triangle determined by the asymptotes of the hy-
perbolax2
4−y2
9− 1 = 0 and the line d : 9x+ 2y − 24 = 0.
3) Find the intersection points between the hyperbolax2
20−y2
5− 1 = 0 and
the line d : 2x− y − 10 = 0.
4) Find the equations of the tangent lines to the hyperbolax2
20−y2
5−1 = 0,
which are orthogonal to d : 4x+ 3y − 7 = 0.
5) Find the equations of the tangent lines to the hyperbolax2
3−y2
5−1 = 0,
passing through P (1,−5).
6) Find the geometric locus of the orthogonal projections of a focus of ahyperbola on the tangent lines to the hyperbola.
5.4 The Parabola
5.4.1 Definition
The parabola is a plane curve defined to be the geometric locus of the pointsin the plane, whose distance to a fixed line d is equal to its distance to a fixedpoint F .
The line d is the director line and the point F is the focus. The distancebetween the focus and the director line is denoted by p and represents theparameter of the parabola.
Consider a Cartesian system of coordinates xOy, in which F
(p
2, 0
)and
d : x = −p
2. If M(x, y) is an arbitrary point of the parabola, then it verifies
|MN | = |MF |,
where N is the orthogonal projection of M on Oy.
116
Figure 5.6:
Thus, the coordinates of a point of the parabola verify√√√√(x+p
2
)2
+ 0 =
√√√√(x− p
2
)2
+ y2 ⇔
⇔
(x+
p
2
)2
=
(x−
p
2
)2
= y2 ⇔
⇔ x2 + px+p2
4= x2 − px+
p2
4+ y2,
and the equation of the parabola is
y2 = 2px. (5.14)
Remark : The equation (5.14) is equivalent to y = ±√
2px, so that theparabola is symmetrical with respect to Ox.
Representing the graph of the function f : [0,∞) → [0,∞) and usingthe symmetry of the curve with respect to Ox, one obtains the graph of theparabola (see Figure (5.7). One has
f ′(x) =p
√2px0
; f ′′(x) = −p
2x√
2x.
117
x 0 ∞f ′(x) | + + + +f(x) 0 ↗ ∞f ′′(x) − − − − −
Figure 5.7:
5.4.2 Intersection of a Parabola and a Line
Let P : y2 = 2px be a parabola, d : y = mx+ n (m 6= 0) be a line and{y2 = 2pxy = mx+ n
be the system determined by their equations. This leads to a second degreeequation
m2x2 + 2(mn− p)x+ n2 = 0,
having the discriminant∆ = 4p(2mn− p) (5.15)
• If ∆ < 0, then the line does not intersect the parabola;
• If ∆ > 0, then there are two intersection points between the line and theparabola;
• If ∆ = 0, then the line is tangent to the parabola and they have a uniqueintersection point.
118
5.4.3 The Tangent to a Parabola
The Tangent of a Given Direction
A line d : y = mx+n (with m 6= 0) is tangent to the parabola P : y2 = 2pxif the discriminant ∆ which appears in (5.15) is zero, i.e. 2mn = p. Then, theequation of the tangent line to P, having the angular coefficient m, is
y = mx+p
2m. (5.16)
The Tangent to a Parabola at a Point of the Parabola
Let P : y2 = 2px be a parabola and P0(x0, y0) be a point of P. Supposethat y0 > 0, so that the point P0 belongs to the graph of the function f :[0,∞) → [0,∞), f(x) =
√2px. The angular coefficient of the tangent at P0
to the curve is
f ′(x0) =p
√2px0
=p
y0.
A similar computation leads to the angular coefficient of the tangent for y0 < 0,
which is stillp
y0.
The equation of the tangent at P0 to P is
y − y0 = f ′(x0)(x− x0),
or, replacing f ′(x0),
y − y0 =p
y0(x− x0) ⇔
⇔ yy0 − y20 = p(x− x0) ⇔
yy0 − 2px0 = p(x− x0),
hence the equation of the tangent is
yy0 = p(x+ x0). (5.17)
5.4.4 Exercises
1) Find the focus and the director line of the parabola y2 − 24x = 0.
2) Find the equation of the parabola having the focus F (−7, 0) and thedirector line x− 7 = 0.
119
3) Find the equation of the tangent line to the parabola y2−8x = 0, parallelto d : 2x+ 2y − 3 = 0.
4) Find the equations of the tangent lines to the parabola y2 − 36x = 0,passing through P (2, 9).
5) Find the equation of the tangent line to y2−4x = 0 at the point P (1, 2).
6) Let P1 : y2 − 2px = 0 and P2 : y2 − 2qx = 0 be two parabolas, with0 < q < p. A mobile tangent to P2 intersects P1 at M1 and M2. Findthe geometric locus of the midpoint of the segment [M1M2].
7) Find the geometric locus of the orthogonal projections of the focus of aparabola on the tangent lines to the parabola.
8) Let A, B and C be three distinct points on the parabola of equationy2 = 2px. The tangent lines at A, B respectively C to the paraboladetermine a triangle A′B′C ′. Prove that the line passing through thecenters of gravity of the triangles ∆ABC and ∆A′B′C ′ is parallel to Ox.
9) Let a, b and c be the tangent lines at three distinct point of a parabolaand ABC the triangle determined by the tangents. Prove that the fo-cus of the parabola belongs to the circumscribed circle of the triangle∆ABC.
5.5 Conics Defined Through a General Equation
5.5.1 Definition
Let be given the Euclidean plane E2 and a Cartesian system of coordinatesxOy, associated to it. Let f : R2 → R be a polynomial function, definedthrough
f(x, y) = a11x2 + 2a12xy + a22y
2 + 2a10x+ 2a20y + a00, (5.18)
with a211 + a2
12 + a222 6= 0.
The set Γ = {P (x, y) : f(x, y) = 0} is called algebraic curve of degree 2,or curve in E2.
Consider the real numbers
∆ =
∣∣∣∣∣∣a11 a12 a10
a12 a22 a20
a10 a20 a00
∣∣∣∣∣∣ ; δ =∣∣∣∣a11 a12
a12 a22
∣∣∣∣ ; I = a11 + a22. (5.19)
120
We shall prove that, after making some changes of coordinates (given, infact, by a rotation and, eventually, a translation of the original system), theequation (5.18) turns into an equation which can be identified to be of one ofthe conics already studied.
Remark : If, in the original system of coordinates, a point P is of coordi-nates P (x, y), then, after a rotation and a translation, the coordinates of itsimage are {
x′ = ax− εby + x0
y′ = bx+ εay + y0, (5.20)
where a2 + b2 = 1 and ε = ±1.Suppose that the point P belongs to the conic Γ. Expressing x and y in
the system of equation (5.20) and replacing them in (5.18), one obtains somepolynomial of degree 2, in variables x′ and y′. It can be verified that thenumbers ∆′, δ′ and I ′, associated to this polynomial, coincide with ∆, δ and,respectively, I.
The numbers ∆, δ and I are the metric invariants of the conic Γ.
Theorem 5.5.1.1. The point M0(x0, y0) is the center of symmetry of theconic Γ if and only if (x0, y0) is a critical point of the function f .
Proof: Let Γ be given by the zeros of f : R2 → R, as in (5.18). After atranslation {
x = x0 + x′
y = y0 + y′,
the equation f(x, y) = 0 becomes
a11x′2 + 2a12x
′y′ + a22y′2 + f ′x(x0, y0)x′ + f ′y(x0, y0)y′ + f(x0, y0) = 0. (5.21)
Suppose that the point M0 is the center of symmetry of the conic Γ. Then,the origin (0, 0) is the center of symmetry in the new system. Hence, bothP (x′, y′) and sO(P )(−x′,−y′) belong to the conic, when P ∈ Γ. Thus,{
a11x′2 + 2a12x
′y′ + a22y′2 + f ′x(x0, y0)x′ + f ′y(x0, y0)y′ + f(x0, y0) = 0
a11x′2 + 2a12x
′y′ + a22y′2 − f ′x(x0, y0)x′ − f ′y(x0, y0)y′ + f(x0, y0) = 0
and f ′x(x0, y0)x′ + f ′y(x0, y0)y′ = 0, for any P (x′, y′) ∈ Γ. Then,{f ′x(x0, y0) = 0f ′y(x0, y0) = 0
,
and (x0, y0) ∈ C(f).
121
Conversely, if (x0, y0) ∈ C(f), then f ′x(x0, y0) = 0 and f ′y(x0, y0) = 0, theequation (5.21) becomes a11x
′2+2a12x′y′+a22y
′2+f(x0, y0) = 0 and, obviously,O(0, 0) is center of symmetry, i.e. M0(x0, y0) is the center of symmetry for Γ.�
5.5.2 Some Classification Algorithms
The centers of symmetry of a conic can give an algorithm to classify theconics. The critical points of f are given by the solutions of the system ofequations {
f ′x(x0, y0) = 0f ′y(x0, y0) = 0
⇐⇒{a11x0 + a12y0 + a10 = 0a12x0 + a22y0 + a20
,
with δ =∣∣∣∣a11 a12
a12 a22
∣∣∣∣. Let r =rank(a11 a12
a12 a22
)and r′ = rank
(a11 a12 a10
a12 a22 a20
).
• If δ 6= 0, then r = r′ = 2 and Γ has a unique center of symmetry. Theconics with this property are: the circle, the ellipse, the hyperbola, a pairof concurrent lines, a point, the empty set. The equation of Γ becomes
a11x′2 + 2a12x
′y′ + a22y′2 +
∆δ
= 0. (5.22)
• If δ = 0 and ∆ 6= 0, then r = 1, r′ = 2 and Γ has no center of symmetry.The parabola has this property.
• If δ = 0 and ∆ = 0, then r = r′ = 1 and Γ has an entire line of centersof symmetry. The conics with this property are: two parallel lines, twoidentical lines, the empty set.
The conics can, also, be classified through the sign of the invariant δ.
• If δ > 0, the conic is said to be of elliptical genus;
• If δ < 0, then the conic is of hyperbolical genus;
• If δ = 0, then the conic is of parabolical genus.
Remark : A particular case of conics is obtained if δ 6= 0 and, in (5.22),a12 = 0 and a11 = a22 = a 6= 0. Then, the equation (5.22) becomes
ax′2 + ay′2 +∆δ
= 0 ⇐⇒ x′2 + y′2 = −∆aδ.
122
• If −∆aδ< 0, then Γ = ∅;
• If −∆aδ
= 0, then Γ = {M0(x0, y0)}; the conic is degenerated to onepoint: its center of symmetry;
• If −∆aδ> 0, then Γ is the circle centered at M0 and of radius
√−
∆aδ
.
5.5.3 Methods of Graphical Representation
Let Γ be the conic defined through (5.18) and ∆, δ and I be the invariantsof Γ, given by (5.19). We saw that, after a translation, the equation of Γ hasbeen reduced to
a11x′2 + 2a12x
′y′ + a22y′2 +
∆δ
= 0.
• If a12 = 0, then one only makes the translation, and Γ has the form
a11x′2 + a22y
′2 +∆δ
= 0.
• If a12 6= 0, then, before making the translation, one makes a rotation(which will cancel the a12) and, after, the translation.
For the latter situation, we present here two methods of representation.
The Eigenvalues Method
Take the matrix A =(a11 a12
a12 a22
), whose determinant is δ. The eigenvalues
of this matrix are the solutions of its characteristic equation:
det (A− λI2) = 0 ⇔∣∣∣∣a11 − λ a12
a12 a22−λ
∣∣∣∣⇔ λ2 − Iλ+ δ = 0. (5.23)
The discriminant of the last equation is
I2−4δ = (a11+a22)2−4(a11a22−a212) = (a11−a22)2+4a2
12 > 0, (since a12 6= 0),
so that the equation (5.23) has two real and distinct solutions, λ1 and λ2.
123
Let a1(u1, v1) and a2(u2, v2) be the eigenvectors associated to the eigen-values λ1 respectively λ2. Their components are, thus, given by the solutionsof the systems{
(a11 − λ1)u1 + a12v1 = 0a12u1 + (a22 − λ1)v1 = 0
, respectively{
(a11 − λ2)u2 + a12v2 = 0a12u2 + (a22 − λ2)v2 = 0
.
Consider the matrix R, whose columns are given by the components ofthe versors e1 and e2 of a1, respectively a2, such that det R = 1 (one might,eventually, replace one of the versors by its opposite, in order to have thedeterminant equal to 1).
The solution of the matrix equation(xy
)= R
(x′
y′
)(which expresses the
rotation) reduces the conic to the following form:
λ1x′2 + λ2y
′2 + 2a′10x′ + 2a′20y
′ + a′00 = 0.
This is equivalent with
λ1
(x′ +
a′10λ1
)2
+ λ2
(y′ +
a′20
λ2
)2
+ a′00 −a′210λ1
− a′220
λ2= 0.
After a translation of equationsx′′ = x′ +
a′10λ1
y′′ = y′ +a′20λ2
,
one obtains the canonical equation of the conic
λ1x′′ + λ2y
′′ + a = 0,
where a = a′00 −a′210
λ1−a′220
λ2.
Example: Let us consider the conic Γ, given by 3x2−4xy−2x+4y−3 = 0.The associated invariants of Γ are
∆ =
∣∣∣∣∣∣3 −2 −1−2 0 2−1 2 −3
∣∣∣∣∣∣ = 8; δ =∣∣∣∣ 3 −2−2 0
∣∣∣∣ = −4; I = 3 + 0 = 3.
124
Since δ 6= 0, then Γ has a unique center of symmetry. Its coordinates aregiven by the solution of the system of equations{
6x− 4y − 2 = 0−4x+ 4 = 0
,
so that the center of symmetry is C(1, 1).Since δ < 0, the conic is of hyperbolical genus.The eigenvalues λ1 and λ2 are the solutions of the equation λ2−3λ−4 = 0,
hence λ1 = −1 and λ2 = 4.Let us determine the eigenvectors associated to λ1 and λ2.
λ1 = −1{
4u1 − 2v1 = 0−2u1 + v1 = 0
⇔ a1(α, 2α), α ∈ R∗ ⇒ e1
(1√
5,
2√
5
)
λ2 = 4{
−u2 − 2v2 = 0−2u2 − 4v2 = 0
⇔ a2(−2β, β), β ∈ R∗ ⇒ e2
(−
2√
5,
1√
5
).
The matrix of the rotation is given by R =
1√
5−
2√
52√
5
1√
5
, and det R = 1.
The equations of the rotation are
(xy
)= R
(x′
y′
)⇐⇒
(xy
)=
1√
5−
2√
52√
5
1√
5
(x′y′)⇐⇒
x =
1√
5x′ −
2√
5y′
y =2√
5x′ +
1√
5y′
.
Replacing in the equation of the conic, one obtains
−x′2 + 4y′2 +6√
5x′ +
8√
5y′ − 3 = 0,
or
−
(x′ −
3√
5
)2
+ 4
(y′ +
1√
5y′
)2
− 2 = 0.
After a translation of equations
x′′ = x′ −
3√
5
y′′ = y′ +1√
5
, the conic is of the form
125
−x′′2 + 4y′′2 − 2 = 0, so that the canonic equation of the given conic is
x′′
2−y′′
12
= −1,
and it is a hyperbola.
Figure 5.8:
The Transformations Method
One can determine the angle of rotation of the coordinate axes. Let Γ be aconic given by (5.18), with a12 6= 0.
Theorem 5.5.3.1. The angle θ of the rotation rθ is given by the equation
(a11 − a22) sin 2θ = 2a12 cos 2θ. (5.24)
Proof: The matrix R is, actually, the matrix of rθ, so that
R =(
cos θ − sin θsin θ cos θ
).
The equations of the rotation are(xy
)= R
(x′
y′
)⇐⇒
(xy
)=(
cos θ − sin θsin θ cos θ
)(x′
y′
)⇐⇒
{x = x′ cos θ − y′ sin θy = x′ sin θ + y′ cos θ
.
126
Replacing in the equation of the conic, one obtains
f(x′ cos θ − y′ sin θ, x′ sin θ + y′ cos θ) = 0 ⇔
⇔ a11(x′ cos θ − y′ sin θ)2 + 2a12(x′ cos θ − y′ sin θ)(x′ sin θ + y′ cos θ)+
+a22(x′ sin θ+y′ cos θ)2+2a10(x′ cos θ−y′ sin θ)+2a20(x′ sin θ+y′ cos θ)+a00 = 0.
The coefficient of x′y′ in this equation is
(a11 − a22) sin 2θ − 2a12 cos 2θ.
But we saw that the effect of the rotation is to cancel this coefficient. Thus
(a11 − a22) sin 2θ − 2a12 cos 2θ = 0. �
Example: Let us take the conic x2 +xy+ y2− 6x− 16 = 0. The invariantsare
∆ =
∣∣∣∣∣∣1 1
2 −312 1 0−3 0 −16
∣∣∣∣∣∣ = −21; δ =∣∣∣∣1 1
212 1
∣∣∣∣ = 34; I = 1 + 1 = 2.
Since δ 6= 0 and δ > 0, then the conic has a unique center of symmetry and isof elliptic genus. The coordinates of the center of symmetry are given by{
2x+ y − 6 = 0x+ 2y = 0
,
so that the center of symmetry is C(4,−2).The angle of rotation is given by
(1− 1) sin 2θ = cos 2θ ⇐⇒ cos 2θ = 0 ⇐⇒ θ =π
4.
The eigenvalues λ1 and λ2 are the solutions of the equation λ2 − 2λ+34
= 0,
hence λ1 =32
and λ2 =12.
The equations of the rotation are
(xy
)= R
(x′
y′
)⇐⇒
(xy
)=
cosπ
4− sin
π
4
sinπ
4cos
π
4
(x′y′
)⇐⇒
x = x′
√2
2− y′
√2
2
y = x′√
22
+ y′√
22
,
127
and the conic becomes
32x′2 +
12y′2 − 3
√2x′ + 3
√2y′ − 16 = 0,
or32(x′ −
√2)2 +
12(y′ + 3
√2)2 − 28 = 0.
After a translation of equations{
x′′ = x′ −√
2y′′ = y′ + 3
√2
, one obtains the reduced
equation of the conicx′′2
563
+y′′2
56= 1,
the conic being an ellipse.
Figure 5.9:
Conclusions
We can put together all the considerations we have made. Let Γ be theconic given by the zeros of the polynomial function f : R2 → R,
f(x, y) = a11x2 + 2a12xy + a22y
2 + 2a10x+ 2a20y + a00,
128
with a211 + a2
12 + a222 6= 0. Let
∆ =
∣∣∣∣∣∣a11 a12 a10
a12 a22 a20
a10 a20 a00
∣∣∣∣∣∣ ; δ =∣∣∣∣a11 a12
a12 a22
∣∣∣∣ ; I = a11 + a22.
be the invariants of Γ.
Conditions The curve Transformations
∆ = 0
δ > 0 Γ = {(x0, y0)}
δ = 0two parallel lines, twoidentical lines, or theempty set
If a12 = 0, one makes a trans-lation;
δ < 0
two concurrent lines; ifI = 0, then the lines areorthogonal.
If a12 6= 0, one makes a ro-tation of angle θ, given by(a11−a22) sin 2θ = 2a12 cos 2θ,and, eventually, a translation.
∆ 6= 0
δ < 0 I∆ < 0 ellipseI∆ > 0 ∅
δ = 0 parabola
δ < 0hyperbola; if I = 0,then the hyperbola isequilateral.
5.5.4 Exercises
1) Find the canonical equation and sketch the graph of the conic
5x2 + 4xy + 8y2 − 32x− 56y + 80 = 0.
2) Find the canonical equation and sketch the graph of the conic
8y2 + 6xy − 12x− 26y + 11 = 0.
3) Find the canonical equation and sketch the graph of the conic
x2 − 4xy + x2 − 6x+ 2y + 1 = 0.
4) Discuss the nature of the conics in the family
x2 + λxy + y2 − 6x− 16 = 0, λ ∈ R.
129
130
Chapter 6
Quadric Surfaces
The quadric surfaces can be seen as the three dimensional analogs of theconics. The set S of the points (x, y, z) ∈ R3, such that g(x, y, z) = 0, whereg ∈ R[X,Y, Z] is a polynomial, is called algebraic surface in R3. If deg(g) = 2,S is said to be a quadric surface in R3. Hence, the points of a quadric surfaceare the zeros of the polynomial function
g(x,y,z)=a11x2+a22y
2+a33z2+2a12xy+2a13xz+2a23yz+2a10x+2a20y+2a30x+a00
with a211 + a2
22 + a233 + a2
12 + a223 + a2
13 > 0.If Oxyz is a Cartesian system of coordinates, the sections determined by
the coordinate planes on a quadric surface are conics, and so are the sectionsdetermined by planes which are parallel to the coordinate planes. By studyingthese sections, one can imagine the shape of a given quadric surface.
The simplest equations for the quadric surfaces are obtained when thesurfaces are situated in certain standard positions relative to the coordinateaxes.
6.1 Ellipsoids
The ellipsoid is the quadric surface given by the equation
E :x2
a2+y2
b2+z2
c2− 1 = 0, a, b, c ∈ R∗+. (6.1)
• The coordinate planes are all planes of symmetry of E since, for anarbitrary point M(x, y, z) ∈ E , its symmetrical points with respect tothese planes, M1(−x, y, z), M2(x,−y, z) and M3(x, y,−z) belong to E ;therefore, the coordinate axes are axes of symmetry for E and the originO is the center of symmetry of the ellipsoid (6.1);
131
• The traces in the coordinates planes are ellipses of equations y2
b2+z2
c2− 1 = 0
x = 0;
x2
a2+z2
c2− 1 = 0
y = 0;
x2
a2+y2
b2− 1 = 0
z = 0;
• The sections with planes parallel to xOy are given by setting z = λ in
(6.1). Then, a section is of equations
x2
a2+y2
b2= 1−
λ2
c2z = λ
.
– If |λ| < c, the section is an ellipse
x2a√
1−λ2
c2
2 +y2b
√1−
λ2
c2
2 = 1
z = λ
;
– If |λ| = c, the intersection is reduced to one (tangency) point(0, 0, λ);
– If |λ| > c, the plane z = λ does not intersect the ellipsoid E .
The sections with planes parallel to xOz or yOz are obtained in a similarway.
The ellipsoid is presented in Figure 6.1.
Figure 6.1:
132
Particular case: If a = b = c = R, one obtains the sphere
x2 + y2 + z2 −R2 = 0
of center O and radius R.
6.2 Hyperboloids of One Sheet
The surface of equation
H1 :x2
a2+y2
b2−z2
c2− 1 = 0, a, b, c ∈ R∗+, (6.2)
is called hyperboloid of one sheet.
• The coordinate planes are planes of symmetry for H1; hence, the co-ordinate axes are axes of symmetry and the origin O is the center ofsymmetry of H1;
• The intersections with the coordinates planes are, respectively, of equa-tions
y2
b2−z2
c2− 1 = 0
x = 0a hyperbola
;
x2
a2−z2
c2− 1 = 0
y = 0a hyperbola
;
x2
a2+y2
b2− 1 = 0
z = 0an ellipse
;
• The intersections with planes parallel to the coordinate planes arey2
b2−z2
c2= 1−
λ2
a2
x = λhyperbolas
;
x2
a2−z2
c2= 1−
λ2
b2y = λ
hyperbolas
;
x2
a2+y2
b2= 1 +
λ2
c2z = λellipses
;
The hyperboloid H1 is presented in Figure 6.2.
Remark : The surface H1 contains two families of lines. One has
x2
a2−z2
c2= 1−
y2
b2⇔
(x
a+z
c
)(x
a−z
c
)=
(1 +
y
b
)(1−
y
b
).
133
Figure 6.2:
The equations of the two families of lines are
dλ :
λ
(x
a+z
c
)= 1 +
y
b
x
a−z
c= λ
(1−
y
b
) , λ ∈ R and d′µ :
µ
(x
a+z
c
)= 1−
y
b
x
a−z
c= µ
(1 +
y
b
) , µ ∈ R.
Through any point on H1 pass two lines, one line from each family.
6.3 Hyperboloids of Two Sheets
The hyperboloid of two sheets is the surface of equation
H2 :x2
a2+y2
b2−z2
c2+ 1 = 0, a, b, c ∈ R∗+. (6.3)
• The coordinate planes are planes of symmetry for H1, the coordinateaxes are axes of symmetry and the origin O is the center of symmetryof H1;
• The intersections with the coordinates planes are, respectively,y2
b2−z2
c2+ 1 = 0
x = 0a hyperbola;
;
x2
a2−z2
c2+ 1 = 0
y = 0a hyperbola
;
x2
a2+y2
b2+ 1 = 0
z = 0the empty set
;
134
• The intersections with planes parallel to the coordinate planes arey2
b2−z2
c2= −1−
λ2
a2
x = λhyperbolas
,
x2
a2−z2
c2= −1−
λ2
b2y = λ
hyperbolas
and
x2
a2+y2
b2= −1 +
λ2
c2z = λ
.
– If |λ| > c, the section is an ellipse;
– If |λ| = c, the intersection reduces to a point (0, 0, λ);
– If |λ| < c, one obtains the empty set.
The hyperboloid of two sheets is presented in Figure 6.3.
Figure 6.3:
6.4 Elliptic Cones
The surface of equation
C :x2
a2+y2
b2−z2
c2= 0, a, b, c ∈ R∗+, (6.4)
is called elliptic cone.
135
• The coordinate planes are planes of symmetry for C, the coordinate axesare axes of symmetry and the origin O is the center of symmetry of C;
• The intersections with the coordinates planes arey2
b2−z2
c2= 0
x = 0two lines
;
x2
a2−z2
c2− 1 = 0
y = 0two lines
;
x2
a2+y2
b2= 0
z = 0the origin O(0, 0, 0)
;
• The intersections with planes parallel to the coordinate planes arey2
b2−z2
c2= −
λ2
a2
x = λhyperbolas
;
x2
a2−z2
c2= −
λ2
b2y = λ
hyperbolas
;
x2
a2+y2
b2=λ2
c2z = λellipses
;
The cone is presented in Figure 6.4.
Figure 6.4:
Remark : If a = b in (6.4), one obtains the equation of a circular cone.
6.5 Elliptic Paraboloids
The surface of equation
Pe :x2
p+y2
q= 2z, p, q ∈ R∗+, (6.5)
is called elliptic paraboloid.
136
• The planes xOz and yOz are planes of symmetry;
• The traces in the coordinate planes arey2
q= 2z
x = 0a parabola
;
x2
p= 2z
y = 0a parabola
;
x2
p+y2
q= 0
z = 0the originO(0, 0, 0)
;
• The intersection with the planes parallel to the coordinate planes arex2
p+y2
q= 2λ
z = λ
,
– If λ > 0, the section is an ellipse;
– If λ = 0, the intersection reduces to the origin;
– If λ < 0, one has the empty set;
and y2
q= 2z −
λ2
px = λ
parabolas
;
x2
p= 2z −
λ2
qy = λ
parabolas
;
The elliptic paraboloid is presented in Figure 6.5.
Figure 6.5:
137
6.6 Hyperbolic Paraboloids
The hyperbolic paraboloid is the surface given by the equation
Ph : −x2
p+y2
q= 2z, p, q ∈ R∗+. (6.6)
• The planes xOz and yOz are planes of symmetry;
• The traces in the coordinate planes are, respectively,y2
q= 2z
x = 0a parabola
;
−x2
p= 2z
y = 0a parabola
;
−x2
p+y2
q= 0
z = 0two lines
;
• The intersection with the planes parallel to the coordinate planes arey2
q= 2z +
λ2
px = λ
parabolas
;
−x2
p= 2z −
λ2
qy = λ
parabolas
;
−x2
p+y2
q= 2λ
z = λhyperbolas
;
The elliptic paraboloid is presented in Figure 6.6.
Figure 6.6:
Remark : The hyperbolic paraboloid contains two families of lines. Since(x√p
+y√q
)(−x√p
+y√q
)= 2z,
138
then the two families are, respectively, of equations
dλ :
−x√p
+y√q
= λ
λ
(x√p
+y√q
)= 2z
, λ ∈ R∗ and d′µ :
x√p
+y√q
= µ
µ
(−x√p
+y√q
)= 2z
, µ ∈ R∗.
6.7 Singular Quadrics
6.7.1 Elliptic Cylinder
The elliptic cylinder is the surface of equation
x2
a2+y2
b2− 1 = 0, a, b > 0, (or
x2
a2+z2
c2− 1 = 0, or
y2
b2+z2
c2− 1 = 0). (6.7)
The graph of the elliptic cylinder is presented in Figure 6.7.
Figure 6.7:
6.7.2 Hyperbolic Cylinder
The hyperbolic cylinder is the surface of equation
x2
a2−y2
b2− 1 = 0, a, b > 0, (or
x2
a2−z2
c2− 1 = 0, or
y2
b2−z2
c2− 1 = 0). (6.8)
The graph of the hyperbolic cylinder is presented in Figure 6.8.
139
Figure 6.8:
6.7.3 Parabolic Cylinder
The parabolic cylinder is the surface of equation
y2 = 2px, p > 0, (or an alternative equation). (6.9)
The graph of the parabolic cylinder is presented in Figure 6.9.
Figure 6.9:
6.7.4 A Pair of Two Planes With Nonempty Intersection
The equationx2
a2−y2
b2= 0, a, b > 0,
140
(or an analogous form) represents the union of two planes π1 :x
a−y
b= 0 and
π2 :x
a+y
b= 0.
6.7.5 A Pair of Two Parallel Planes
The equationx2 − a2 = 0, a > 0,
(or a similar form) represents the union of two parallel planes π1 : x − a = 0and π2 : x+ a = 0.
6.7.6 Two Identical Planes
The equation x2 = 0, for instance, represents the union of two identicalplanes.
6.7.7 The Line
The equationx2
a2+y2
b2= 0 represents a line in 3-space (in fact, the Oz axes).
6.7.8 The Point
The point P0(x0, y0, z0) is given by the solution of the equation
(x− x0)2
a2+
(y − y0)2
b2+
(z − z0)2
c2= 0.
6.7.9 The Empty Set
It can be seen as the quadric given by the equation
x2
a2+y2
b2+z2
c2= −1.
6.8 The Classification Algorithm
Let Q be the quadric surface given by the zeros of the polynomial
g(x,y,z)=a11x2+a22y
2+a33z2+2a12xy+2a13xz+2a23yz+2a10x+2a20y+2a30x+a00
with a211 + a2
22 + a233 + a2
12 + a223 + a2
13 > 0.
141
Consider the matrices
A =
a11 a12 a13
a12 a22 a23
a13 a23 a33
and A′ =
a00 a10 a20 a30
a10 a11 a12 a13
a20 a12 a22 a23
a30 a13 a23 a33
.
Denote by r =rank A, r′ =rank A′, δ = detA and ∆ = detA′. Let P (λ)be the characteristic polynomial of the matrix A, i.e.
P (λ) = det(A− λI3) =
∣∣∣∣∣∣a11 − λ a12 a13
a12 a22 − λ a23
a13 a23 a33 − λ
∣∣∣∣∣∣ .Since A is symmetric, then P (λ) has three real roots λ1, λ2 and λ3. These arethe eigenvalues of A. Let i be the number of negative eigenvalues (obviously,0 ≤ i ≤ 3).
Now, the characteristic polynomial is
P (λ) = I0 − I1λ+ I2λ2 − λ3,
whereI0 = δ = λ1λ2λ3,
I1 =∣∣∣∣a11 a12
a12 a22
∣∣∣∣+ ∣∣∣∣a11 a13
a13 a33
∣∣∣∣+ ∣∣∣∣a22 a23
a23 a33
∣∣∣∣ = λ1λ2 + λ1λ3 + λ2λ3,
I2 = a11 + a22 + a33 = Tr(A) = λ1 + λ2 + λ3.
Consider
P0(λ) =
∣∣∣∣∣∣∣∣a00 a10 a20 a30
a10 a11 − λ a12 a13
a20 a12 a22 − λ a23
a30 a13 a23 a33 − λ
∣∣∣∣∣∣∣∣ = k0 − k1λ+ k2λ2 − λ3,
wherek0 = ∆,
k1 =
∣∣∣∣∣∣a00 a10 a20
a10 a11 a12
a20 a12 a22
∣∣∣∣∣∣+∣∣∣∣∣∣a00 a10 a30
a10 a11 a13
a30 a13 a33
∣∣∣∣∣∣+∣∣∣∣∣∣a00 a20 a30
a20 a22 a23
a30 a23 a33
∣∣∣∣∣∣ ,k2 =
∣∣∣∣a00 a30
a30 a33
∣∣∣∣+ ∣∣∣∣a00 a20
a20 a22
∣∣∣∣+ ∣∣∣∣a00 a10
a10 a11
∣∣∣∣ .142
Theorem 6.8.1. Let Q be the quadric surface given by g(x, y, z) = 0, whereg is a polynomial of degree 2. There exists a Cartesian system of coordinatesin which the quadric has one of the following forms:
1) If δ 6= 0, then Q : λ1x′2 + λ2y
′2 + λ3z′2 +
∆δ
= 0;
2) If δ = 0, ∆ 6= 0, I1 6= 0, then Q : λ1x′2 + λ2y
′2 ±
√−
∆I1z′ = 0;
3) If δ = 0, ∆ = 0, I1 6= 0, then Q : λ1x′2 + λ2y
′2 +k1
I1= 0;
4) If δ = 0, I1 = 0, k1 6= 0, then Q : I2x′2 + 2
√−k1
I2y′ = 0;
5) If δ = 0, I1 = 0, k1 = 0, then Q : I2x′2 +k2
I2= 0.
143
Conclusions
r′ r i Quadric Canonical equation Remarks Remarks∆ > 0 ∅
3 ∆ < 0 ellipsoid
∆ > 0hyperboloidof onesheet
λ1x′2 + λ2y
′2 + λ3z′2 +
∆δ
= 0
non-singularquadrics3
2∆ < 0
hyperboloidof twosheets
4
2 elliptic paraboloidλ1x
′2 + λ2y′2 ±
√−
∆I1z′ = 02 1 hyperbolic paraboloid
3 a pointλ1x
′2 + λ2y′2 + λ3z
′2 = 03 2 elliptic cone
32 2
k1I1 < 0ellipticcylinder
λ1x′2 + λ2y
′2 +k1
I1= 0
non-degene-ratedquadrics
k1I1 > 0 ∅1 hyperbolic cylinder
1 1 parabolic cylinder I2x′2 + 2
√−k1
I2y′ = 0 singular
quadrics
2
22 a line
λ1x′2 + λ2y
′21 two planes
1 1k2 > 0 ∅
I2x′2 +
k2
I2= 0
k2 < 0 twoparallelplanes
degene-ratedquadrics
1 1 1 a double plane x2 = 0
6.9 Exercises
1) Sketch the graph of the ellipsoidx2
4+y2
16+z2
9= 1.
2) Sketch the graph of the hyperboloid of one sheet x2 + y2 −z2
4= 1.
3) Sketch the graph of the hyperboloid of two sheets x2 +y2
4− z2 = −1.
144
4) Sketch the graph of the elliptic cone z2 = x2 +y2
4.
5) Sketch the graph of the elliptic paraboloid z =x2
4+y2
9.
6) Sketch the graph of the hyperbolic paraboloid z =x2
4−y2
9.
7) Sketch the graph of the surface z = 1− x2 − y2.
8) Sketch the graph of the surface 4x2 + 4y2 + z2 + 8y − 4z = −4.
9) Show that the lines
x = 3 + ty = 2 + tz = 5 + 2t
and
x = 3 + ty = 2− tz = 5 + 10t
are completely
contained into the hyperbolic paraboloid z = x2 − y2.
10) Name and sketch the surface
a) z = (x+ 2)2 + (y − 3)2 − 9;
b) 4x2 − y2 + 16(z − 2)2 = 10;
c) 4x2 + y2 − z2 + 8x− 2y + 4z = 0;
d) 9x2 + y2 + 4z2 − 18x+ 2y + 16z − 10 = 0.
145
146
Chapter 7
Generated Surfaces
Consider the 3-dimensional Euclidean space E3, together with a Cartesiansystem of coordinates Oxyz. Generally, the set
S = {M(x, y, z) : F (x, y, z) = 0},
where F : D ⊆ R3 → R is a real function andD is a domain, is called surface ofimplicit equation F (x, y, z) = 0 (the quadric surfaces, defined in the previouschapter for F a polynomial of degree two, are such of surfaces). On the otherhand, the set
S1 = {M(x, y, z) : x = x(u, v), y = y(u, v), z = z(u, v)},
where x, y, z : D1 ⊆ R2 → R, is a parameterized surface, of parametric equa-tions
x = x(u, v)y = y(u, v)z = z(u, v)
, (u, v) ∈ D1.
The intersection between two surfaces is a curve in 3-space (remember, forinstance, that the intersection between a quadric surface and a plane is a conicsection, hence the conics are plane curves). Then, the set
C = {M(x, y, z) : F (x, y, z) = 0, G(x, y, z) = 0},
where F,G : D ⊆ R3 → R, is the curve of implicit equations{F (x, y, z) = 0G(x, y, z) = 0
.
147
As before, one can parameterize the curve. The set
C1 = {M(x, y, z) : x = x(t), y = y(t), z = z(t)},
where x, y, z : I ⊆ R → R and I is open, is called parameterized curve ofparametric equations
x = x(t)y = y(t)z = z(t)
, t ∈ I.
Let be given a family of curves, depending on one single parameter λ,
Cλ :{F1(x, y, z;λ) = 0F2(x, y, z;λ) = 0
.
In general, the family Cλ does not cover the entire space. By eliminating theparameter λ between the two equations of the family, one obtains the equationof the surface generated by the family of curves.
Suppose now that the family of curves depends on two parameters λ, µ,
Cλ,µ :{F1(x, y, z;λ, µ) = 0F2(x, y, z;λ, µ) = 0
,
and that the parameters are related through ϕ(λ, µ) = 0 (one can chooseonly the sub-family corresponding to such λ and µ). If it can be obtainedan equation which does not depend on the parameters (by eliminating theparameters between the three equations), then the set of all the points whichverify it is called surface generated by the family (or the sub-family) of curves.
7.1 Cylindrical Surfaces
The surface generated by a variable line (the generatrix ), which remainsparallel to a fixed line d and intersects a given curve C, is called cylindricalsurface. The curve C is called the director curve of the cylindrical surface.
Theorem 7.1.1. The cylindrical surface, with the generatrix parallel to theline d, where
d :{π1 = 0π2 = 0
,
and having the director curve C, where
C :{F1(x, y, z) = 0F2(x, y, z) = 0
,
148
Figure 7.1:
(suppose that d and C are not coplanar), is characterized by an equation of theform
ϕ(π1, π2) = 0. (7.1)
Proof: An arbitrary line, which is parallel to d :{π1(x, y, z) = 0π2(x, y, z) = 0
, will
be of equations
dλ,µ :{π1(x, y, z) = λπ2(x, y, z) = µ
.
Of course, not every line from the family dλ,µ intersects the curve C. Thishappens only when the system of equations
F1(x, y, z) = 0F2(x, y, z) = 0π1(x, y, z) = λπ2(x, y, z) = µ
is compatible. By eliminating x, y and z between the four equations of thesystem, one obtains a compatibility condition for the parameters λ and µ,
ϕ(λ, µ) = 0.
The equation of the surface can be determined now from the systemπ1(x, y, z) = λπ2(x, y, z) = µϕ(λ, µ) = 0
,
and it is immediate that ϕ(π1, π2) = 0. �
149
Remark : Any equation of the form (7.1), where π1 and π2 are linear func-tion of x, y and z, represents a cylindrical surface, having the generatrices
parallel to d :{π1 = 0π2 = 0
.
Example: Let us find the equation of the cylindrical surface having thegeneratrices parallel to
d :{x+ y = 0z = 0
and the director curve given by
C :{x2 − 2y2 − z = 0
x− 1 = 0.
The equations of the generatrices d are
dλ,µ :{x+ y = λz = µ
.
They must intersect the curve C, i.e. the systemx2 − 2y2 − z = 0
x− 1 = 0x+ y = λz = µ
has to be compatible. A solution of the system can be obtained using thethree last equations
x = 1y = λ− 1z = µ
and, replacing in the first one, one obtains the compatibility condition
2(λ− 1)2 + µ− 1 = 0.
The equation of the surface is obtained by eliminating the parameters inx+ y = λz = µ
2(λ− 1)2 + µ− 1 = 0.
Then,2(x+ y − 1)2 + x− 1 = 0.
150
7.2 Conical Surfaces
The surface generated by a variable line, which passes through a fixed pointV and intersects a given curve C, is called conical surface. The point V iscalled the vertex of the surface and the curve C director curve.
Figure 7.2:
Theorem 7.2.1. The conical surface, of vertex V (x0, y0, z0) and directorcurve
C :{F1(x, y, z) = 0F2(x, y, z) = 0
,
(suppose that V and C are not coplanar), is characterized by an equation ofthe form
ϕ
(x− x0
z − z0,y − y0
z − z0
)= 0. (7.2)
Proof: The equations of an arbitrary line through V (x0, y0, z0) are
dλµ :{x− x0 = λ(z − z0)y − y0 = µ(z − z0)
.
A generatrix has to intersect the curve C, hence the system of equationsx− x0 = λ(z − z0)y − y0 = µ(z − z0)F1(x, y, z) = 0F2(x, y, z) = 0
151
must be compatible. This happens for some values of the parameters λ andµ, which verify a compatibility condition
ϕ(λ, µ),
obtained by eliminating x, y and z in the the previous system of equations.In these conditions, the surface is generated and its equation rises from the
system x− x0 = λ(z − z0)y − y0 = µ(z − z0)
ϕ(λ, µ) = 0.
It follows that
ϕ
(x− x0
z − z0,y − y0
z − z0
)= 0.. �
Remark : If ϕ is an algebraic function, then the equation (7.2) can bewritten in the form φ(x−x0, y− y0, z− z0) = 0, where φ is homogeneous withrespect to x− x0, y − y0 and z − z0.
If ϕ is algebraic and V is the origin of the system of coordinates, thenthe equation of the conical surface is φ(x, y, z) = 0, with φ a homogeneouspolynomial. Conversely, an algebraic homogeneous equation in x, y and zrepresents a conical surface with the vertex at the origin.
Example: Let us determine the equation of the conical surface, having thevertex V (1, 1, 1) and the director curve
C :{
(x2 + y2)2 − xy = 0z = 0
.
The family of lines passing through V has the equations
dλµ :{x− 1 = λ(z − 1)y − 1 = µ(z − 1)
.
The system of equations (x2 + y2)2 − xy = 0
z = 0x− 1 = λ(z − 1)y − 1 = µ(z − 1)
152
must be compatible. A solution isx = 1− λy = 1− µz = 0
,
and, replaced in the first equation of the system, gives the compatibility con-dition
[(1− λ)2 + (1− µ)2]2 − (1− λ)(1− µ) = 0.
The equation of the conical surface is obtained by eliminating the parametersλ and µ in
x− 1 = λ(z − 1)y − 1 = µ(z − 1)
((1− λ)2 + (1− µ)2)2 − (1− λ)(1− µ) = 0.
Expressing λ =x− 1z − 1
and µ =y − 1z − 1
and replacing in the compatibility con-
dition, one obtains(z − x
z − 1
)2
+
(z − y
z − 1
)22
−
(z − x
z − 1
)(z − y
z − 1
)= 0,
or[(z − x)2 + (z − y)2]2 − (z − x)(z − y)(z − 1)2 = 0.
7.3 Conoidal Surfaces
The surface generated by a variable line, which intersects a given line d anda given curve C, and remains parallel to a given plane π, is called conoidalsurface. The curve C is the director curve and the plane π is the director planeof the conoidal surface.
Theorem 7.3.1. The conoidal surface whose generatrix intersects the line
d :{π1 = 0π2 = 0
and the curve
C :{F1(x, y, z) = 0F2(x, y, z) = 0
153
Figure 7.3:
and has the director plane π = 0, (suppose that π is not parallel to d and thatC is not contained into π), is characterized by an equation of the form
ϕ
(π,π1
π2
)= 0. (7.3)
Proof: An arbitrary generatrix of the conoidal surface is contained into aplane parallel to π and, on the other hand, comes from the bundle of planescontaining d. Then, the equations of a generatrix are
dλµ :{
π = λπ1 = µπ2
.
Again, the generatrix must intersect the director curve, hence the system ofequations
π = λπ1 = µπ2
F1(x, y, z) = 0F2(x, y, z) = 0
has to be compatible. This leads to a compatibility condition
ϕ(λ, µ) = 0,
and the equation of the conoidal surface is obtained fromπ = λ
π1 = µπ2
ϕ(λ, µ) = 0.
154
By expressing λ and µ, one obtains (7.3). �
Example: Let us find the equation of the conoidal surface, whose genera-trices are parallel to xOy and intersect Oz and the curve{
y2 − 2z + 2 = 0x2 − 2z + 1 = 0
.
The equations of xOy and Oz are, respectively,
xOy : z = 0, and Oz :{x = 0z = 0
,
so that the equations of the generatrix are
dλ,µ :{x = λyz = µ
.
From the compatibility of the system of equationsx = λyz = µ
y2 − 2z + 2 = 0x2 − 2z + 1 = 0
,
one obtains the compatibility condition
2λ2µ− 2λ2 − 2µ+ 1 = 0,
and, replacing λ =y
xand µ = z, the equation of the conoidal surface is
2x2z − 2y2z − 2x2 + y2 = 0.
7.4 Revolution Surfaces
The surface generated after the rotation of a given curve C around a givenline d is said to be a revolution surface.
Theorem 7.4.1. The equation of the surface generated by the curve
C :{F1(x, y, z) = 0F2(x, y, z) = 0
,
155
Figure 7.4:
in its rotation around the line
d :x− x0
p=y − y0
q=z − z0
r,
is of the form
ϕ((x− x0)2 + (y − y0)2 + (z − z0)2, px+ qy + rz) = 0. (7.4)
Proof: An arbitrary point on the curve C will describe, in its rotationaround d, a circle situated into a plane orthogonal on d and having the centeron the line d. This circle can be seen as the intersection between a sphere,having the center on d and of variable radius, and a plane, orthogonal on d,so that its equations are
Cλ,µ :{
(x− x0)2 + (y − y0)2 + (z − z0)2 = λpx+ qy + rz = µ
.
The circle has to intersect the curve C, therefore the systemF1(x, y, z) = 0F2(x, y, z) = 0
(x− x0)2 + (y − y0)2 + (z − z0)2 = λpx+ qy + rz = µ
must be compatible. One obtains the compatibility condition
ϕ(λ, µ) = 0,
which, after replacing the parameters, gives the equation of the surface (7.4).�
156
Example: Let us determine the equation of the torus (the surface generatedby a circle C, which turns around an exterior line, lying in the plane of thecircle). Choose the system of coordinates such that Oz is the line d and Ox isthe orthogonal line on d, passing through the center of C. Let r be the radiusof the circle and (0, a, 0) the coordinates of its center. Since the line is exteriorto the circle, then a > r > 0.
Figure 7.5:
In this system of coordinates, the equations of the circle and of the lineare, respectively,
C :{
(y − a)2 + z2 = r2
x = 0and d :
{x = 0z = 0
.
The equations of the family of circles generating the surface are
Cλ,µ :{x2 + y2 + z2 = λ
z = µ.
The system of equations x2 + y2 + z2 = λ
z = µx = 0
(y − a)2 + z2 = r2
must be compatible. Choose the first three equations in order to obtain asolution of the system
x = 0y = ±
√λ− µ2
z = µ
.
157
Replacing in the remained equation, one obtains the compatibility condition
(±√λ− µ2 − a)2 + µ2 = r2.
The equation of the torus is
(±√x2 + y2 − a)2 + z2 = r2,
or(x2 + y2 + z2 + a2 − r2)2 = 4a2(x2 + y2).
7.5 Exercises
1) Find the equation of the cylindrical surface, having the circle
C :{x2 + y2 − a2
z = 0
as director curve, and the generatrices parallel to d : x = y = z.
2) Find the equation of the cylindrical surface, generated by a variable lineof direction (1, 2,−1) and of director curve
C :{
x+ y + z = 0x2 + y2 + z2 = 4
.
3) Find the equation of the cylindrical surface, having the director curve
C :{x2 + y2 − z = 0
x = 2z
and the generatrices orthogonal on the plane containing C.
4) Find the equation of the conical surface, with the vertex V (0,−a, 0) andthe director curve
C :{x2 + y2 + z2 = 4
y + z = 2.
5) A circular disk of center (1, 0, 2) and radius 1 is parallel to the planeyOz. Supposing that there is a light source at the point P (0, 0, 3), findthe shadow that the disk darts on the plane xOy.
158
6) Find the geometric locus of the lines passing through the origin O(0, 0, 0)and tangent to the sphere
(x− 5)2 + (y + 1)2 + z2 − 16 = 0.
7) Find the equation of the surface generated by a line which intersects Oz,the line
d :{
x− z = 0x+ 2y − 3 = 0
and stays parallel to xOy.
8) Find the equation of the conoidal surface, generated by a line whichintersects a line d, a circle situated in a plane parallel to d and whichstays parallel to a plane orthogonal on d (the Willis’ conoid).
9) Find the equation of the surface generated by the rotation of the lined : x = y = z around the axis Oz.
10) Find the equation of the surface generated by the rotation of a linearound another line.
11) Find the equation of the surface generated by the rotation of the hyper-bola
H :
z = 0x2
a2−y2
b2− 1 = 0
around the axis Oy.
159
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