Polar Coordinates P T P T O T

ضل والتكامل المتقدمالتفا

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Polar Coordinates جامعة القادسية / كلية التربية / قسم الرياضيات / الصف الثاني

Polar Coordinates

Definition : The polar coordinate system is a two-dimensional coordinate system in which

each point P on a plane is identified by the ordered pair (r, ). The number measures the

angle between the positive x-axis and a ray that goes through the point P, as shown in figure,

the number r measures the distance from the origin to the point P, and r, are called polar

coordinates.

Remark : We extend the meaning of polar coordinates (r, ) to the case in which r is

negative by agreeing that the points (-r, ) and (r, ) lie in the same line through O and at the

same distance |r| from O, but on opposite sides of O. If r > 0, the point (r, ) lies in the same

quadrant as ; if r < 0, it lies in the quadrant on the opposite side of the pole.

Example : Plot the points whose polar coordinates are given:

(a) (1, 5π/4) (b) (2, 3π) (c) (2,−2π/3) (d) (−3, 3π/4)

Solution:


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Remark : In the Cartesian coordinate system every point has only one representation, but in

the polar coordinate system each point has many representations. For instance, the point (1,

5π/4) in the Example above could be written as (1,−3π/4) or (1, 13π/4) or (−1, π/4).

Example : Find all the polar coordinates of the point P(2, π/6).

Solution : We sketch the initial ray of the coordinate system, draw the ray from the origin

that makes an angle of π/6 radians with the initial ray, and mark the point (2, π/6). We then

find the angles for the other coordinate pairs of P in which r = 2 and r = −2.

For r = 2, the complete list of angles is

π/6 , π/6 ± 2π , π/6 ± 4π , π/6 ± 6π , …

For r = −2, the angles are

5π/6 , 5π/6 ± 2π , 5π/6 ± 4π , 5π/6 ± 6π , …

The corresponding coordinate pairs of P are

(2, π/6 + 2nπ) , n = 0, ±1, ±2, . . .

and

(-2, π/6 + 2nπ) , n = 0, ±1, ±2, . . .

When n = 0, the formulas give (2, π/6) and (−2,−5π/6).

When n = 1, they give (2, 13π/6) and (−2, 7π/6), and so on.

Relationship between Polar and Rectangular Coordinates

The connection between polar and Cartesian coordinates can be seen from the figure below

and described by the following formulas:

x = r cos , y = r sin

r2 = x2 + y2 , tan = y/x


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Example :

(a) Convert the point (2, π/3) from polar to Cartesian coordinates.

(b) Represent the point with Cartesian coordinates (1,−1) in terms of polar coordinates.

Solution:

(a) We have:

x = r cos = 2 cos π/3 = 2 · 1/2 = 1

y = r sin = 2 sin π/3 = 2. √3 /2 = √3

Therefore, the point is (1, √3) in Cartesian coordinates.

(b) If we choose r to be positive, then

𝑟 = √𝑥2 + 𝑦2 = √12 + −12 = √2

tan =y/x=-1

Since the point (1,−1) lies in the fourth quadrant, we can choose = −π/4 or = 7π/4. Thus

one possible answer is (√2,−π/4); another is (√2, 7π/4).

Polar Equations

Example : Find the equation of x = 1 in polar coordinates.

Solution: We use the formula x = r cos .

x = 1 → r cos = 1 → r = sec

Example : Find the equation of x2 = 4y in polar coordinates.

Solution: We use the formulas x = r cos and y = r sin.

x2 = 4y → (r cos)

2 = 4 r sin → r2

cos2 = 4 r sin

→ r = sec tan

Example : Find the equation of the circle (x − 1/2)2 + y

2 = 1/4 in polar coordinates.

Solution:

(x − 1/2)2 + y

2 = 1/4 → (r cos − 1/2)

2 + r

2 sin

2 = 1/4 →

r2 cos

2 − r cos + 1/4 + r2 sin

2 = 1/4 → r2 (cos

2 + sin2 ) − r cos = 0

→ r2 − r cos = 0 → r (r − cos) = 0

If r≠0 then r − cos =0 → r = cos

Example : Find the equation of the line y = 3x + 2 in polar coordinates.

Solution: H.W.


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Polar Curves

The graph of a polar equation r = f(); or more generally F(r , ) = 0; consists of all points

P that have at least one polar representation (r , ) whose coordinates satisfy the equation.

Example : Express the polar equation in rectangular coordinates. If possible, determine the

graph of the equation from its rectangular form.

(a) r = 5 sec (b) r = 2 sin (c) r = 2 + 2 cos (d) r=2

Solution:

(a) and (b) and (d) H.W

(c) r = 2 + 2 cos → (r = 2 + 2 cos ) × r

→ r2 = 2r + 2r cos → x

2 + y

2 = 2r +2x

→ x2 + y

2 - 2x = 2r → (x2

+ y2 - 2x)

2 = 4r

2 → (x2

+ y2 - 2x)

2 = 4(x

2 + y

2)

Exercises

1- Plot these polar coordinate points on one graph: (2, π/3), (−3, π/2), (−2,−π/4), (1/2, π),

(1, 4π/3), (0, 3π/2).

2- Find an equation in polar coordinates to the following given equations in rectangular

coordinates :

i) y = 3x ii) y = sin x iii) y = 3x2 -2x iv) y = x

2 + y

2

3- Sketch the curve to the following :

i) r = cos ii) r = sin( + π /4) iii) r = cot csc iv) r2 = −2 sec csc

4- Find an equation in rectangular coordinates to the following given equation in polar

coordinates:

5- i) r = sin(3) ii) r = sin2 iii) r = sec csc iv) r = tan


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Tangents to Polar Curves

To find a tangent line to a polar curve r = f() we regard as a parameter and write its

parametric equations as

x = r cos = f() cos , y = r sin = f() sin

Then, using the method for finding slopes of parametric curves and the product rule, we have

𝑑𝑦

𝑑𝑥=

𝑑𝑦

𝑑𝜃

𝑑𝑥

𝑑𝜃

=

𝑑𝑟

𝑑𝜃 𝑠𝑖𝑛𝜃 + 𝑟 𝑐𝑜𝑠𝜃

𝑑𝑟

𝑑𝜃 𝑐𝑜𝑠𝜃 – 𝑟 𝑠𝑖𝑛𝜃

Example : For the curve r = 1 + sin , find the slope of the tangent line when = π/3.

𝑑𝑦

𝑑𝑥=

𝑑𝑦𝑑𝜃𝑑𝑥𝑑𝜃

=

𝑑𝑟𝑑𝜃

𝑠𝑖𝑛𝜃 + 𝑟 𝑐𝑜𝑠𝜃

𝑑𝑟𝑑𝜃

𝑐𝑜𝑠𝜃 – 𝑟 𝑠𝑖𝑛𝜃=

𝑐𝑜𝑠𝜃 𝑠𝑖𝑛𝜃 + (1 + 𝑠𝑖𝑛𝜃)𝑐𝑜𝑠𝜃

𝑐𝑜𝑠𝜃 𝑐𝑜𝑠𝜃 − (1 + 𝑠𝑖𝑛𝜃)𝑠𝑖𝑛𝜃∴

𝑑𝑦

𝑑𝑥

=𝑐𝑜𝑠𝜃 (1 + 2 𝑠𝑖𝑛𝜃)

1 − 2 𝑠𝑖𝑛2𝜃 − 𝑠𝑖𝑛𝜃

The slope of the tangent at the point = π/3 is

𝑑𝑦

𝑑𝑥|

𝜃=𝜋/3=

cos (𝜋

3)(1+2 sin(

𝜋

3))

(1+sin(𝜋

3))(1−2 sin(

𝜋

3))

=1/2(1+√3)

(1+√3

2)(1−√3)

= −1


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Conic Sections (Conics) جامعة القادسية / كلية التربية / قسم الرياضيات / الصف الثاني

Conic Sections (Conics)

In general, the equations of the general quadratic form are equations of the form

Ax2 + Bxy + Cy

2 + Dx + Ey + F = 0. The graphs of these equations represent the conic

sections (circles, parabolas, ellipses, and hyperbolas), except for some few cases.

Classification of conics by general quadratic equation

Conics can be classified (or distinguished) by computing the discriminant, B2-4AC, of the

general quadratic equation, Ax2 + Bxy + Cy

2 + Dx + Ey + F = 0. We will not be studying any

conics with the Bxy term, so we will always assume that B = 0. If B ≠ 0, then the conic is

rotated such that its major (and minor) axis is no longer parallel to one of the coordinate axes.

1. If B2 - 4AC > 0, then the conic is a hyperbola. This is equivalent to showing that A and

C have opposite signs.

2. If B2 - 4AC = 0, then the conic is a parabola. This is equivalent to showing that either A

or C is equal to zero.

3. If B2 - 4AC < 0, then the conic is either an ellipse or a circle. This is equivalent to

showing that A and C have the same sign. Circles are distinguished from ellipses when

A = C.

Example : Identify each of the following conics:

x2 + 2y

2 - 4x + 6y - 1= 0 (ellipse)

2x2 + 2y

2 - 4x + 6y - 1= 0 (circle)

x2 - 2y

2 - 4x + 6y -1= 0 (hyperbola)

x2 - 4x + 6y -1= 0 (parabola)

Exercise Classify each of the following conics:

1. x2 − 2xy − 3y

2 + x − 1 = 0

2. 2x2 + xy − y

2 − 2x + 3y = 0

3. 4x2 − y+3 = 0

4. −x2 − xy − y

2 + 3x = 0

5. x2 + 2x − y − 3 = 0

6. 8x2 + 12xy + 17y

2 − 20 = 0

7. x2 + xy − 1 = 0

8. 4x2 − y

2 − 4y = 0

9. 6x2 + 9y

2 − 24x − 54y+51 = 0


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The parabola

A parabola is the set of all points in a plane whose distance from a fixed point (called the

focus ) is equal to its distance from a fixed line (called the directrix ). The line passing

through the focus and perpendicular to the directrix is called its axis of symmetry, and the

point where the parabola intersects its axis of symmetry is called its vertex.

We will use this definition to derive an equation of a parabola with vertex at the origin, focus

at F(0, p), p>0, and directrix y = - p, and let P(x,y) is any point on the parabola. Using the

distance formula between two points, we can write √(𝑥 − 0)2 + (𝑦 − 𝑝)2 = 𝑑1. The

distance between P and Q is just the vertical distance y + p = d2, then by the definition of a

parabola, we have d1 = d2 or

→ √(𝑥 − 0)2 + (𝑦 − 𝑝)2 = 𝑦 + 𝑝. Squaring both sides, we obtain

→ (x - 0)2 + (y - p)

2 = (y + p)

2 → x

2 + y

2 - 2py + p

2 = y

2 + 2 py + p

2

→ x2 = 4 py.

Then an equation of the parabola with focus (0,p), p>0, and directrix y=-p is x2= 4py.

Remark : There are another three cases of the parabola (Check with the graphic):

i. The parabola with focus (0,p), p<0, and directrix y=-p is x2= 4py.

ii. The parabola with focus (p,0), p>0, and directrix x=-p is y2= 4px.

iii. The parabola with focus (p,0), p<0, and directrix x=-p is y2= 4px.

Example 1: Find the focus and the directrix and sketch the graph of the parabola :


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1) y2+10x=0 2)y

2 = 8x. 3) y

2 − x+3 = 0 4)4x

2=-y 5)4y+x

2=0

Example 2 : Find an equation for the parabola with vertex (0,0) , focus(0,-2).

Translations (Shifted Conics)

We can translate the graph of a parabola h units horizontally by replacing x with (x - h) and k

units vertically by replacing y with (y - k). This leads to four standard equations of parabolas,

one for each orientation.

Example 1: Find the focus, directrix, and vertex and then sketch the graph of the parabola

1) y2+2y+8x+17 = 0 2)x

2-2x=-8y+7 3)x

2+4x=4y 4)y

2+4x=8

5) x2+2x-4y-3=0 6)y

2+x+y=0 7)x

2-8y=0

Example 2 : Find an equation and the focus and sketch the parabolas :

1)V(2,0) , the directrix is y-axis 2)V(-3,1) , the directrix is x=1.

3)V(-2,-2) , the directrix is y=-3 4) V(1,-2) , the directrix is x-axis.

Example 2 : Find an equation and the directrix and sketch the parabolas :

1)V(0,0) , F(0,2) 2)V(0,3) , F(-1,3) 3)V(-2,3) , F(-2,4) 4)V(1,-3) , F(1,0)

Ellipses

An ellipse is the set of all points in the plane, the sum of whose distances from two fixed

points F1 and F2 are called the foci (plural of focus) is constant.

In order to obtain the simplest equation for an ellipse, we place the foci on the x-axis at the

points (-c,0) and (c,0) so that the origin is halfway between the foci. Let the sum of the

distances from a point on the ellipse to the foci be 2a>0. Then P(x,y) is a point on the ellipse

when |PF1| + |PF2| = 2a

That is ….

(a2-c

2)x

2+a

2y

2=a

2(a

2-c

2)

From the tringle F1F2P we have that 2c<2a so c<a

and therefore a2-c

2>0, now let b

2=a

2-c

2.

Then the equation of the ellipse becomes

b2x

2+a

2y

2=a

2b

2 →

𝑥2

𝑎2+

𝑦2

𝑏2= 1

since b2=a

2-c

2 < a

2 it follows that b<a.


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Remark 1 If we sitting y=0 then we have x2/a

2=1 or x

2=a

2 so x=±a, then the x-intercepts

points are (a,0) and (-a,0) , called the vertices of ellipse, and the line segment joining the

vertices is called the major axis. To find the y-intercepts points we set x=0 and obtain y2=b

2

so y=±b i.e. the points are (0,b) and (0,-b) and the line segment joining between them is

called the minor axis.

Remark 2 : The another cases of the ellipse (Check with the graphic) is 𝑥2

𝑏2+

𝑦2

𝑎2= 1 where

a≥b>0 and the foci (0,±c) and vertices (0,±a).

Remark 3 : If the foci coincide then c=0 and , a=b and ellipse becomes a circle with radius

r=a=b.

Example Sketch the ellipse 𝑥2

4+

𝑦2

9= 1

Example Sketch the graph of 9x2 16y2

= 144.

Translations (Shifted Conics)

If we translate the center of an ellipse to the point (h, k), then we obtain the standard forms: (𝑥−ℎ)2

𝑎2+

(𝑦−𝑘)2

𝑏2= 1 and

(𝑥−ℎ)2

𝑏2+

(𝑦−𝑘)2

𝑎2= 1

The foci, vertices, and endpoints are translated similarly.


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Hyperbolas

A hyperbola is the set of all points in the plane, the difference of whose distances from two

fixed points F1 and F2 (called the foci) is a positive constant. That is, for any point P(x,y) on

the hyperbola curve we have | |PF1|-|PF2| |=2a and a>0.

In order to obtain the simplest equation for an hyperbola, we place the foci on the x-axis at

the points (-c,0) and (c,0) so that the origin is halfway between the foci. Let the difference

between the distances from a point on the hyperbola to the foci be 2a>0. Then P(x,y) is a

point on the hyperbola when | |PF1| - |PF2| | = 2a.

Then the equation of the hyperbola of this case is 𝑥2

𝑎2−

𝑦2

𝑏2= 1 (check)

Where c2=a

2+b

2, therefore c>a.

Remarks:

The x-intercepts are ±a and the points (a,0) and (-a,0) are the vertices of the hyperbola.

There is no y-intercept. (why?)

Unlike with the ellipse a is not always greater than b.

The hyperbola is symmetric with respect to both axes.

From the equation to the hyperbola we have

x2/a

2 - y

2/b

2 = 1 → x

2/a

2 = 1 + y

2/b

2 ≥ 1 → x

2/a

2 ≥ 1

→ x2 ≥ a

2 → |x| ≥ a → x ≥ a or x ≤ a

This means that the hyperbola consists of two parts, called its branches.

When we draw a hyperbola it is useful to first draw its asymptotes, which are the lines

y=(b/a)x and y=-(b/a)x.

The another case of hyperbola when the foci are on the y-axis, then we have

y2/a

2 - x

2/b

2 = 1

where the foci (0,±c) , c2=a

2+b

2 , vertices (0,±a), and asymptotes are y=±(a/b)x.

We shift conics by taken the properties with replacing x and y by x-h and y-k.


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Example 1 : Find the foci, vertices and asymptotes and sketch the graph to the following

hyperbolas :

1)9x2-16y

2=144 2) y

2 - 4x

2=1 3)2y

2-3x

2=1 4)12x

2-4y

2=48

5)x2/25- y

2/16=1 6) 9x

2 – 4y

2 – 72x + 8y + 176 = 0

7) 2y2 – 3x

2 - 4y + 12x + 8 = 0 8)16x

2 – 9y

2 + 64x - 90y = 305

Example 2 : Find the equation of the hyperbola with the following properties and sketch the

graph :

1)V(0,±1) and asymptote y=2x 2)V(±4,0) and asymptote y=3/4x

3)V1(4,4) , V2(4,-2) and asymptote y-1=3/2(x-4)

4)F1(2,2) , F2(6,2) and asymptotes y=x-2 and y=6-x

5)F1(2,-2), F2(2,8), V1(2,0) and V2(2,6)

6)F1(1,3), F2(7,3), V1(2,3) and V2(6,3)

التفاضل والتكامل المتقدم

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The Triple Perpendicular Coordinates جامعة القادسية / كلية التربية / قسم الرياضيات / الصف الثاني

Three Dimensional Spaces

Let’s first take a look at Two Dimensional Cartesian (rectangular) Coordinate System or

the Euclidean plane , it is consists of two perpendicular coordinate axes x-axis and y-axis with

appropriate units in both axes. Any point P on the plane is associated with an ordered pair of

real numbers (x , y) in a unique way as follows: x = intersection of the vertical line passing

through P and x-axis (called x-coordinate), y = intersection of the horizontal line passing

through P and y-axis (called y-coordinate). The origin is the point (0; 0).

For instance, as in figure 1, the point P(1 , 2) is formed as the intersection of a vertical line

across x-axis at x=1 and a horizontal line across y-axis at y=2.

The plane is also called xy-plane. The xy-plane is also known as the set R

2 = {(x , y) : x and

y are real numbers}, the 2 represents the number of dimensions of the plane. The axes, which

are collectively referred to as the coordinate axes, divided the plane into four quadrants, as in

figure 2.

We now generalize these concepts to three-dimensional Cartesian (rectangular) coordinate

system, or xyz-space or the Euclidean space by introducing a third axis called z-axis that is

perpendicular to the xy-plane at the origin. In this space, a point P is represented by an

ordered triple (x,y,z) that consists of three numbers, an x-coordinate, a y-coordinate, and a z-

coordinate. As in the two-dimensional xy-plane, these coordinates indicate the signed

distance along the coordinate axes, the x-axis, y-axis and z-axis, respectively, from the origin,

denoted by O, which has coordinates (0,0,0). The xyz-space is also known as the set R

3 = {(x,y,z) : x,y and z are real numbers}.

The projection of the point (x,y,z) onto the xy-plane is the point (x,y,0). Similarly, the

projection of this point onto the yz-plane is the point (0,y,z), and the projection of this point

onto the xz-plane is the point (x,0,z), as in figure 3.

Example : What are the orthogonal projection of (2,-4, 3) on xy, yz and zx-planes?


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Graphing points P(a,b,c) works much like in two dimensions, except we also include depth as

following (figure 4):

1. We move a steps along the x-axis, i.e. we arrive the point (a,0,0).

2. We, then, move b steps in the y-direction, i.e. we arrive the point (a,b,0).

3. Finally, we move c steps in the z-direction, which finishes our path to (a,b,c).

Note that we can take this path in any order. We can move in the y-direction, then the z-

direction, and finish with movement in the x-direction.

Unlike two-dimensional space, which consists of a single plane(the xy-plane) three-

dimensional space contains infinitely many planes, just as two-dimensional space consists of

infinitely many lines. Three planes are of particular importance (figure 5):

1. The xy-plane, which contains the x- axes and y-axes, i.e. the set {(x,y,z)R3: z = 0}.

2. The yz-plane, which contains the y- axes and z-axes, i.e. the set {(x,y,z)R3 : x = 0}.

3. The xz-plane, which contains the x- axes and z-axes, i.e. the set {(x,y,z)R3 : y = 0}.


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Just as the x-axis and y-axis divide the xy-plane into four quadrants, these three planes

divide xyz-space into eight octants. Within each octant, all x-coordinates have the same sign,

as do all y-coordinates, and all z-coordinates. In particular, the first octant is the octant in

which all three coordinates are positive (figure 6).

Figure 6

Example 1 : Plot and determine the octant of the following points:

A(-2,3,0) , B(1,0,1), C(0,-4,1) , D(-3,-1,0)

Example 2 : Which of the axis contains the following points:

A(0,0,1) , B(0,1,0) , C(2,0,0)


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Distance between two points When P1(x1,y1,z1) and P2(x2,y2,z2) are distinct points in xyz-space, the distance between the

two points is

|𝑃1𝑃2| = √(𝑥1 − 𝑥2)2 + (𝑦1 − 𝑦2)

2 + (𝑧1 − 𝑧2)2

Example : Find the distance between the points A and B to the following :

1- A(2,4,-4) , B(3,-3,2) 2- A(1,3,2) , B(4,3,1) 3- A(0,2,3) , B(3,2,3)

The sphere A sphere is defined as the set of all points in three-dimensional Euclidean space R

3 that are

located at a distance r (the radius) from a given point (the center (a,b,c)). Twice the radius is

called the diameter. The Cartesian equation of a sphere centered at the point (a,b,c) with

radius r is given by

(x - a)2 + (y - b)

2 + (z - c)

2 = r

2

Example 1 : Find the equation of the sphere with center at Q and radius r to the following:

1- Q(2,3,0) , r=2 2- Q(1,-3,2) , r=3 3- Q(3,1,2) , r=1

Example 2 : Find the equation of the sphere that passes through the point P(1,2,4) and has

center at Q(2,3,1).

Example 3 : Determine if the equation is a sphere

1- x2+ y

2 + z

2 = 6x + 4y + 10z

2- 2x2 + 2y

2 + 2z

2 + 4y - 2z+2=0

3- x2 + y

2 + 2z

2 + 2x - z+3y=0

Example 4 : Find the center and the radius of the sphere x2+y

2+z

2+4x-2y+6z-2=0.

Example 5 : Consider the point P such that the distance from P to A(-1,5,3) is twice the

distance from P to B(6,2,-2) . Show that the set of all such points is a sphere. Find the center

and radius of the sphere.


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The Vectors in Space جامعة القادسية / كلية التربية / قسم الرياضيات / الصف الثاني

The Vectors in Space

The directed line segments are used to describe or represent the forces, accelerations,

velocities and displacements in space. This directed line segments are called vectors. The

vectors have both a magnitude and a direction.

Definition : A (nonzero) vector is a directed line segment drawn from a point P (called its

initial point) to a point Q (called its terminal point), with P and Q being distinct points. The

vector is denoted by 𝑃𝑄⃗⃗⃗⃗ ⃗. Its magnitude is the length of the line segment, denoted by ||𝑃𝑄⃗⃗⃗⃗ ⃗||, and its direction is the same as that of the directed line segment. The zero vector is just a

point, and it is denoted by 0.

Definition : Two nonzero vectors are equal if they have the same magnitude and the same

direction. Any vector with zero magnitude is equal to the zero vector.

By this definition, vectors with the same magnitude and direction but with different initial

points would be equal. For example, in Figure 8 the vectors u, v and w all have the same

magnitude but we have u=w and u≠v (why?).

So we can see that there are an infinite number of vectors for a given magnitude and

direction, those vectors all being equal and differing only by their initial and terminal points.

Remark : Unless otherwise indicated, when speaking of the vector with a given magnitude

and direction, we will mean the one whose initial point is at the origin of the coordinate

system.


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Example : Let v be the vector in R3 whose initial point is at the origin and whose terminal

point is P(3,4,5). Though the point P(3,4,5) and the vector v are different objects, it is

convenient to write v = (3,4,5). When doing this, it is understood that the initial point of v is

at the origin (0,0,0) and the terminal point is P(3,4,5).

Remark : The point-vector correspondence provides an easy way to check if two vectors are

equal, without having to determine their magnitude and direction. Since the vectors starting at

the origin, we are now only seeing if the terminal points are the same. To get new vectors

starting at the origin, we translate each vector to start at the origin by subtracting the

coordinates of the original initial point from the original terminal point. The resulting point

will be the terminal point of the “new” vector whose initial point is the origin. Do this for

each original vector then compare.

Example : Consider the vectors 𝑃𝑄⃗⃗⃗⃗ ⃗ and 𝑅𝑆⃗⃗⃗⃗ ⃗ in R3, where P=(2,1,5), Q= (3,5,7), R=(1,−3,−2)

and S=(2,1,0). Does 𝑃𝑄⃗⃗⃗⃗ ⃗ = 𝑅𝑆⃗⃗⃗⃗ ⃗?

Solution: The vector 𝑃𝑄⃗⃗⃗⃗ ⃗ is equal to the vector v with initial point (0,0,0) and terminal point

Q−P=(3,5,7)−(2,1,5)=(3−2,5−1,7−5) = (1,4,2).

Similarly, 𝑅𝑆⃗⃗⃗⃗ ⃗ is equal to the vector w with initial point (0,0,0) and terminal point

S−R=(2,1,0)−(1,−3,−2) = (2−1,1−(−3),0−(−2)) = (1,4,2).

So 𝑃𝑄⃗⃗⃗⃗ ⃗=v = (1,4,2) and 𝑅𝑆⃗⃗⃗⃗ ⃗=w= (1,4,2).

∴𝑃𝑄⃗⃗⃗⃗ ⃗ = 𝑅𝑆⃗⃗⃗⃗ ⃗


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Theorem : For a vector v = (a,b,c) in R3, the magnitude (length) of v is:

||𝑣|| = √𝑎2 + 𝑏2 + 𝑐2

Example : Calculate the magnitudes (lengths) of the following vectors:

(a)v=(2,−1,4) (b) v = (2,−1,0) (c) v = (3,2,−2) (d) v = (0,0,1) (e) v = (6,4,−4)

Vector Algebra

Definition : For a scalar k and a nonzero vector v, the scalar multiple of v by k, denoted by kv,

is the vector whose magnitude is |k| ||v|| i.e. ||kv||=|k| ||v|| points in the same direction as v if k

> 0, points in the opposite direction as v if k < 0, and is the zero vector 0 if k = 0. For the zero

vector 0, we define k0 = 0 for any scalar k.

Remark : Two vectors v and w are parallel (denoted by v||w) if one is a scalar multiple of the

other.

Remark : In special case if v(a,b,c), whose the initial point is the origin, then kv=(ka,kb,kc).

Definition : The sum of vectors v and w, denoted by v+w, is obtained by translating w so that

its initial point is at the terminal point of v, the initial point of v+w is the initial point of v,

and its terminal point is the new terminal point of w.

In general, since the scalar multiple –v = −1v is a well-defined vector, we can define vector

subtraction as follows v−w= v+(−w).


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Remark : In special case if v(a1,b1,c1) and w(a2,b2,c2), where the initial point is the origin,

then v+w =(a1+a2, b1+b2, c1+c2).

Example : Find kv, v+w, v-w for the following:

1) v(2,4,-1), w(3,2,4), k=2. 2) v(-3,0,1), w(0,1,-2), k=-3.

Theorem : For any vectors u(a1,b1,c1), v(a2,b2,c2), w(a3,b3,c3), and scalars k, l, we have

i. v+w=w+v Commutative Law

ii. u+(v+w) = (u+v)+w Associative Law

iii. v+0 = v = 0+v Additive Identity

iv. v+(−v) = 0 Additive Inverse

v. k(lv) = (kl)v Associative Law

vi. k(v+w) = kv+kw Distributive Law

vii. (k+l)v = kv+lv Distributive Law

Proof :


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A unit vector

A unit vector is a vector with magnitude 1. Notice that for any nonzero vector v, the vector 𝑣

||𝑣|| is a unit vector which points in the same direction as v, since

1

||𝑣|| > 0 and ||

𝑣

||𝑣||||=

||𝑣||

||𝑣|| =1.

Dividing a nonzero vector v by ||v|| is often called normalizing v.

Example : Which of the following vectors are unit vector. Normalizing the vector if it not

unit vector :

1) v(1,3,-1) 2) v(2

√6 ,

1

√6 ,

−1

√6 ) 3) v(2,-1,3) 4) v(

1

√2 ,

1

2 ,

1

2 )

The basis vectors :

Let v(a,b,c) be any vector in R3 then by vector algebra we can write this vector as following:

(a,b,c)=(a,0,0) + (0,b,0) + (0,0,c)

Also we have

(a,0,0)=a(1,0,0) & (0,b,0)=b(0,1,0) & (0,0,c)=c(0,0,1)

Then we can write any vector v(a,b,c) as a unique scalar combination of the vectors

{(1,0,0),(0,1,0),(0,0,1)} as following :

(a,b,c) = a(1,0,0) + b(0,1,0) + c(0,0,1)

For example (3,4,-2)=3(1,0,0)+4(0,1,0)-2(0,0,1) & (5,0,2)=5(1,0,0) + 2(0,0,1)

Therefore the set of vectors { (1,0,0) , (0,1,0) , (0,0,1) } is a basis of the vectors in R3, which

is called the basis vectors.

Now if we set i=(1,0,0) , j=(0,1,0) and k=(0,0,1) then the set { i , j , k } is another

representation for above basis of the vectors in R3 and any vector v(a,b,c) we can write as

following

(a,b,c) = ai + bj + ck

For example (3,4,-2)=3i+4j-2k & (5,0,2)=5i+2k


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Example 1 : Let v=(−1,5,−2) and w=(3,1,1).

(a) Find v−w. (b) Find v+w. (c) Find 𝑣

||𝑣|| (d) Find ||1/2(v-w)||

(e) Find ||1/2(v+w)|| (f) Find −2v+4w (g) Find v−2w.

(h) Find the vector u such that u+v+w= i.

(i) Find the vector u such that u+v+w= 2j+k.

(j) Is there a scalar m such that m(v+2w) = k? If so, find it.

Example 2 : Let v=(−2,3,1) and w=(2,-3,0), what is the relation between

1) ||v-w|| and ||v||-||w||

2) ||v+w|| and ||v||+||w||


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Dot Product

Definition : Let v=(v1,v2,v3) and w=(w1,w2,w3) be vectors in R3. The dot product of v and w,

denoted by v.w, is given by:

v.w = v1w1 +v2w2 +v3w3

Similarly, if we represented the vectors v = v1i+v2j+v3k and w= w1i+w2j+w3k the dot product

is still v.w = v1w1 +v2w2 +v3w3.

Remark : Notice that the dot product of two vectors is a scalar, not a vector. So the

associative law does not hold for the dot product of vectors. Why?

Example : Find the dot product between v and w to the following vectors:

1)v(2,5,-1) , w(1,0,4) 2) v(-3,2,1) , w(2,4,-5)

Example 2 : Let w(-2,1,0), r(-1,-1,1) and s(5,2,-3) be vectors, find the vector v such that

v.w=v.r=v.s=0.

Documents

Polar Coordinates P T P T O T