New 3 Dimensional Geometry

7/29/2019 New 3 Dimensional Geometry

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Page - [1]ist Of Formulae By OP Gupta www.theOPGupta.WordPress.com

Important Terms, Definitions & Formulae

01. Distance formula: The distance between two points 1 1 1A , , y z and 2 2 2B , , x y z is given by the expression

2 2 2

2 1 2 1 2 1AB units x x y y z z .

02. Section formula: The coordinates of a point Q which divides the line joining the points 1 1 1A , , x y z and

2 2 2B , , y z in the ratio :m n

a) internally, are2 1 2 1 2 1

, ,

mx nx my ny mz nz

m n m n m n

b) externally i.e. , internally in the ratio ( ) : ( )m n , are 2 1 2 1 2 1, ,

mx nx my ny mz nz

m n m n m n.

03. Direction Cosines of a Line: If A and B are two points on a given line L then, the direction cosines of vectors AB and BA arethe direction cosines (d.c.’s) of line L. Thus if α, β, γ are the direction-angles which the line Lmakes with the positive direction of x, y, z- axes respectively then, its d.c.’s are cosα, cosβ, cosγ (See Fig.1). If direction of line L is reversed, the direction angles are replaced by their supplementsi.e. , α, β, γ and so are the d.c.’s i.e., the direction cosines become

cosα, cosβ, cos γ . So, a line in space has two sets of d.c.’s viz. cos α, cosβ, cos γ .

The d.c.’s are generally denoted by , ,l m n . Also 2 2 2 1 l m n and so, we can deduce that2 2 2cos α cos β cos γ 1 . Also 2 2 2sin α sin β sin γ 2 .

The d.c.’s of a line joining the points 1 1 1A , , x y z and 2 2 2B , , y z are

2 1 2 1 2 1, ,AB AB AB

x y y z z ; where AB is the distance between points A and B i.e. ,

2 2 22 1 2 1 2 1AB x y y z z .

In order to obtain the d.c.’s of a line which does not pass through the origin, we draw a linethrough the origin and parallel to the given line. As parallel lines have same set of the d.c.’s, sothe d.c.’s of given line can be obtained by taking the d.c.’s of the parallel line through origin.

In a unit vector, the coefficient of ˆˆ ˆ, ,i j k are called d.c.’s. For example in ˆˆ â li mj nk Thed.c.’s are , ,l m n .

04. Direction Ratios of a Line: Any three numbers , ,a b c (say) which are proportional to d.c.’s i.e., , ,l m n of a line are called the

direction ratios (d.r.’s) of the line. Thus, , , a l b m c n for any R 0 .

Consider, 1 (say)

l m na b c

3 Dimensional Geometry Formulae For By OP Gupta [Indira Award Winner, +91-9650 350 480]

RECAPITULATION, DIRECTION COSINES,

DIRECTION RATIOS & EQUATION OF LINES



List Of Formulae for Class XII By OP Gupta (Electronics & Communications Engineering)


, ,

a b cl m n

2 2 2

1

a b c 2 2 2Using 1 l m n

2 2 2 a b c

Therefore,2 2 2 2 2 2 2 2 2

, ,

a b cl m n

a b c a b c a b c.

The d.c.’s of a line joining the points 1 1 1A , , x y z and 2 2 2B , , x y z are

2 1 2 1 2 1, , x y y z z or 1 2 1 2 1 2, , x x y y z z .

Direction ratios are sometimes called as Direction Numbers as well.

For a line if , ,a b c are its d.r.’s then, , ,ka kb kc ; 0k is also a set of its d.r.’s. So, for a linethere are infinitely many sets of the direction ratios.

05. Relation between the direction cosines of a line: Consider a line L with d.c’s , ,l m n . Draw a line passing through the origin and P , , x y z and

parallel to the given line L. From P draw a perpendicular PA on the X-axis. Suppose OP r . (SeeFig.2)

Now in OAP we have,OA

cosαOP x

lr r

.

Similarly we can obtain mr and z nr .

Therefore, 2 2 2 2 2 2 2 y z r l m n .

But we know that 2 2 2 2 x y z r .

Hence , 2 2 2 1 l m n .

06. Equation of a line in space passing through a given point and parallel to a given vector:Consider the line L is passing through the given point 1 1 1A , , x y z with the position vector a , b

is the given vector with d.r.’s , ,a b c and r is the position vector of any arbitrary point P , , x y z on the line. See Fig.3.Thus 1 1 1

ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ÔA , OP ,a x i y j z k r xi yj zk b ai bj ck

.

a) Vector equation of a line: As the line L is parallel to given vector b and points A and P arelying on the line so, AP is parallel to the b .

AP ,b

where R , set of real nos.i.e.

r a b

r a b .

This is the vector equation of line .

b) Parametric equations: If d.r.’s of the line are , ,a b c then by using r a b we get,

1 1 1ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ y z x y z a b ci j k i j k i j k .

Now, as we equate the coefficients of ˆˆ ˆ, ,i j k we get the Parametric equations of line given as,

1 1 1, , x x a y y b z z c .

Coordinates of any point on the line considered here are 1 1 1

, , x a y b z c .

c) Cartesian equation of a line: If we eliminate the parameter from the Parametric equations of a line, we get the Cartesian equation of line as



MATHEMATICS – List Of Formulae for Class XII By OP Gupta (+91-9650350480)


1 1 1 x y y z z a b c

.

If , ,l m n are the d.c.’s of the line then, Cartesian equation of line becomes

1 1 1 x y y z z l m n

.

The Cartesian equation of line is also called the symmetrical equation or one point form of line . In the symmetrical form the coefficient of x, y, z are unity i.e., 1.

Note that b is parallel to the line L. So they both have the same d.r.’s.

07. Equation of a Line passing through two given points:Consider the two given points as 1 1 1A , , x y z and 2 2 2B , , y z with position vectors a and b

respectively. Also assume r as the position vector of any arbitrary point P , , y z on the line L passing through A and B. See Fig.4.

Thus 1 1 1 2 2 2ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ÔA , OB , OPa x i y j z k b x i y j z k r xi yj zk

.

a) Vector equation of a line: Since the points A, B and P all lie on the same line which meansthey are all collinear points .

Further it means, AP r a and AB b a are collinear vectors, i.e.,AP AB

r a b a

,r a b a where R .

This is the vector equation of line .

b) Cartesian equation of a line: By using the vector equation of the line r a b a we get,

1 1 1 2 1 2 1 2 1ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ x y z x y z x x y y z z i j k i j k i j k

On equating the coefficients of ˆˆ ˆ, ,i j k we get,

1 2 1 1 2 1 1 2 1, , x x x x y y y y z z z z ....( )i

On eliminating we have,1 1 1

2 1 2 1 2 1

x x y y z z x y y z z

.

This is the Cartesian equation of line .

c) Parametric equations: By using (i), we get 1 2 1 1 2 1 1 2 1, , x x x x y y y y z z z z .

These are called the Parametric equations of line .

08. Angle between two Lines: a) When d.r.’s or d.c.’s of the two lines are given:

Consider two lines 1L and 2L with d.r.’s as 1 1 1, ,a b c and 2 2 2, ,a b c ; d.c.’s as 1 1 1, ,l m n and

2 2 2, ,l m n . Consider 1 1 1 1ˆˆ ˆb a i b j c k and 2 2 2 2

ˆˆ ˆb a i b j c k . These vectors 1b and 2b are

parallel to the given lines 1L and 2L . So in order to find the angle between the Lines 1L and

2L , we need to get the angle between the vectors 1b and 2b . Consider the Fig.5.

So the acute angle θ between the vectors 1b and 2b (and hence lines 1L and 2L ) can be

obtained as,1 2 1 2. cosθb b b b





Thus, 1 2 1 2 1 2

2 2 2 2 2 21 1 1 2 2 2

cosθa a b b c c

a b c a b c

.

1 2 1 2 1 2Also, in terms of d.c.’s : cosθ l l m m n n .

2 2 2

1 2 2 1 1 2 2 1 1 2 2 1

2 2 2 2 2 2

1 1 1 2 2 2

Sine of angle is given as: sinθa b a b b c b c c a c a

a b c a b c

.

b) When vector equations of two lines are given:

Consider vector equations of lines 1L and 2L as 1 1 1r a b

and 2 2 2r a b

respectively.Then, the acute angle θ between the two lines is given by the relation

1 2

1 2

.cosθ

b b

b b

.

c) When Cartesian equations of two lines are given:Consider the lines 1L and 2L in Cartesian form as,

1 1 1 2 2 21 2

1 1 1 2 2 2

L : , L : x x y y z z x x y y z z

a b c a b c

.

Then the acute angle θ between the lines 1L and 2L can be obtained by,

1 2 1 2 1 2

2 2 2 2 2 21 1 1 2 2 2

cosθa a b b c c

a b c a b c

.

For two perpendicular lines: 1 2 1 2 1 2 0a a b b c c ; 1 2 1 2 1 2 0l l m m n n .

For two parallel lines: 1 1 1

2 2 2

a b ca b c

; 1 1 1

2 2 2

l m nl m n

.

09. Shortest Distance between two Lines: If two lines are in the same plane i.e. they are coplanar, they will intersect each other if they arenon-parallel. Hence shortest distance between them is zero . If the lines are parallel then the shortestdistance between them will be the perpendicular distance between the lines i.e. the length of the

perpendicular drawn from a point on one line onto the other line. Adding to this discussion, inspace, there are lines which are neither intersecting nor parallel . In fact, such pair of lines are non-coplanar and are called the skew lines .

Skew lines: Two straight lines in space which are neither parallel nor intersecting are known asthe skew lines . They lie in different planes and are non-coplanar.

Shortest Distance: There exists unique line perpendicular to each of the skew lines 1L and

2L , this line is known as line of shortest distance (S.D.).

a) Shortest distance between two Skew lines: When lines are in vector form:

Consider the two skew lines 1 1 1r a b

and 2 2 2r a b

. Assume that A and B are two

points on the lines 1L and 2L with position vectors 1a and 2a respectively. Let us assumethat the Shortest Distance between the two lines is PQ d . See Fig.6.





Now PQ is perpendicular to both the lines 1L and 2L . That means PQ is perpendicular to

both 1b and 2b . But we know that 1 2b b

is perpendicular to both 1b and 2b . So we can

deduce that PQ is in the direction of 1 2b b

.

Let the unit vector in the direction of 1 2b b

is n .

So, 1 2

1 2

ˆb b

nb b

ˆPQ PQ. n

1 2

1 2

PQ PQb b

b b

The shortest distance PQ is basically the projection of AB on PQ .

ˆ. . PQ AB.i e , n Then the S.D. between them is given as follow,

1 2 2 1

1 2

.PQ

b b a ad

b b

.

When the lines are in Cartesian form:Consider the two skew lines as,

1 1 1 2 2 21 2

1 1 1 2 2 2

L : , L : x y y z z x x y y z z

a b c a b c

.

Then the S.D. between them is given as follow

2 1 2 1 2 1

1 1 1

2 2 22 2 2

1 2 2 1 1 2 2 1 1 2 2 1

x x y y z z

a b c

a b cd a b a b b c b c c a c a

.

b) Shortest distance between two parallel lines: If the lines are parallel then they are coplanar i.e. they lie in the same plane.

Consider the two parallel lines as 1 1L : r a b and 2 2L : r a b

. Assume that A and B

are two points on the lines 1L and 2L with position vectors 1a and 2a respectively. Also

assume that the lines are parallel to b . Let AB makes angle θ with the line 1L . So the angle between the AB and b will be θ . See Fig.7.

Draw 1BP L . Now BP represents the perpendicular distance between 1L and 2L .In APB , we have BP = ABsin θ ...( )i

Now consider ÂB AB sin( θ)b b n

ÂB AB sinθ [ 1b b n

AB AB sinθb b

AB BP [ ( )b b By using i





ABBP

b

b

.

Assume that the Shortest Distance between the two lines is BP d .

Then the S.D. between them is given as follow,

2 1b a ad

b

.

Note that the S.D. between two parallel lines in the Cartesian form can be obtained by simply

replacing 2 1 2 1 2 1 2 1ˆˆ ˆ( ) ( ) ( )a a x x i y y j z z k

and ˆˆ ˆb ai bj ck in theexpression obtained above for the Vector form.

Important Terms, Definitions & Formulae

01. Plane and its equation: A plane is a surface such that if any two points are taken on it, the line segment joining them liescompletely on the surface. Plane is symbolized by the Greek letter .

a) Equation of plane in Normal unit vector form: Consider a plane at distance d from the origin such that ON is the normal from the origin to

plane and n is a unit vector along ON . Then ÔN dn if ON d . Consider r be the position vector of any arbitrary point P , , x y z on the plane. See Fig.8.

Vector form of the equation of plane: Since P lies on the plane so NP is perpendicular to

the vector ON . That implies NP. ON 0

ˆ ˆ. 0r dn dn

ˆ ˆ. 0 [ 0r dn n d

ˆ ˆ ˆ. . 0r n dn n

ˆ.r n d ˆ ˆ[ . 1n n This is the vector equation of the plane.

Cartesian form of the equation of plane: If , ,l m n are d.c.’s of the normal n to the given plane. Then by using ˆ.r n d we get,

.ˆ ˆˆ ˆ ˆ ˆ x y z l m ni j k i j k d

lx my nz d .This is the Cartesian equation of the plane.

Also if , ,a b c are the d.r.’s of the normal n to the plane then, the Cartesian equation of plane becomes ax by cz d .

b) Equation of plane Perpendicular to a given vector and passing through a given point: Assume that the plane passes through a point 1 1 1A , , x y z with the position vector a and is

perpendicular to the vector m with d.r.’s as A,B,C ˆˆ Â B Cm i j k .

Also consider P , , x y z as any arbitrary point on the plane with position vector as r .Consider the Fig.9.

PLANE & ITS EQUATION IN VARIOUS FORMS





Vector form of the equation of plane: As AP lies in the plane and m is perpendicular tothe plane. So AP is perpendicular to m .

AP. 0m

. 0r a m .

This is the Vector equation of the plane .

The above obtained equation of plane can also be expressed as . .r m a m

.

Cartesian form of the equation of plane: As 1 1 1 ˆˆ ÂP x i y y j z z k , so byusing . 0r a m

we get,

1 1 1ˆ ˆˆ ˆ ˆ ˆ. A B C 0 x x i y y j z z k i j k

1 1 1A B C 0 x x y y z z .This is the Cartesian equation of the plane .

c) Equation of plane passing through three non- collinear points:Assume that the plane contains three non- collinear points 1 1 1R , , x y z , 2 2 2S , , x y z and

3 3 3T , , y z with the position vectors as a , b and c respectively. Let P , , x y z be anyarbitrary point in the plane whose position vector is r .

Vector form of the equation of plane: As RS and RT are in the plane, soRS RT m

( ) say will be perpendicular to the plane containing the points R , S and T. Also

since r is position vector of P which lies in the plane, therefore RP m . See Fig.10.RP. 0m

. RS RT 0 [ RS RTr a m

. 0r a b a c a

.

This is the Vector equation of the plane .

Cartesian form of the equation of plane: The position vector of RP , RS and RT is given as,

1 1 1 1 1 1

2 2 2 1 1 1 2 1 2 1 2 1

3 3 3 1 1 1 3 1 3 1 3 1

ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆRP

ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆRS

ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆRT

xi yj zk x i y j z k x x i y y j z z k

i y j z k x i y j z k x x i y y j z z k

x i y j z k x i y j z k x x i y y j z z k

Substituting these in the above obtained vector equation of plane , we get

1 1 1

2 1 2 1 2 1

3 1 3 1 3 1

0

x x y y z z

x x y y z z

x x y y z z

.

This is the Cartesian equation of the plane .

d) Intercept form of the equation of plane: Consider the equation of plane A B C D 0, x y z D 0 and the plane makes intercepts

, ,a b c on , , x y z -axes respectively. This implies that the plane meets , , x y z -axes at ,0,0a ,

0, ,0b , 0,0, c respectively. See Fig.11.

Therefore,D

A. B.0 C.0 D 0 A ,a + + + = =a





DA.0 B. C.0 D 0 B+ b + + = =

b and

DA.0 B.0 C. D 0 C+ + c + = =

c .

Substituting these values in A B C D 0, x y z , we get 1 x y z a b c

.

This is the equation of plane in intercept form .

NOTE Equation of XY-plane : 0,

Equation of YZ-plane : 0,

Equation of ZX-plane : 0.

z

x

y

02. Equation of plane passing through the intersection of two given planes: The intersection of two planes say 1 and 2 is always a straight line. For instance, we canvisualize the intersection of y - plane and z - plane to form x- axis .

a)

Vector equation of the plane: Consider two planes 1 1 1: .r m d

and 2 2 2: .r m d

. So if h is the position vector of any arbitrary point on the line of intersection of 1 and 2 then, itmust satisfy both the equations of planes i.e.,

1 1.h m d and 2 2.h m d

1 1. 0h m d and 2 2. 0h m d .Therefore for all R (set of all real nos.) , we get

1 1 2 2. . 0h m d h m d

1 2 1 2.h m m d d

As the obtained equation is of the form .r m d

(Note that in .r m d

, d is not the perpendicular distance of plane from the origin. Rather d is perpendicular distance from theorigin in ˆ.r n d .)

So it represents a plane 3 (say).

Hence, the required plane is: 1 2 1 2.r m m d d .

This is the Vector equation of plane .

b) Cartesian equation of the plane:Assume 1 1 1 1

ˆˆ Â B Cm i j k , 2 2 2 2ˆˆ Â B Cm i j k and ˆˆ ˆr xi yj zk .

Then by using 1 2 1 2.r m m d d , we get

1 2 1 2 1 2 1 2A A B B C C y z d d

i.e. , 1 1 1 1 2 2 2 2A B C A B C 0 x y z d x y z d .This is the Cartesian equation of plane .(You can visualize the situation discussed here in the Fig.12)

03. Co- planarity of two Lines: Assume the given lines are 1 1 1L : r a b

and 2 2 2L : r a b such that 1L passes through

1 1 1A , , x y z with position vector 1a and is parallel to 1b with d.r.’s 1 1 1, ,a b c . Also 2L passes

through 2 2 2B , , x y z with position vector 2a and is parallel to 2b with the d.r.’s 2 2 2, ,a b c .

a) Vector form of co- planarity of lines:





We know 2 1AB a a . Now the lines 1L and 2L are coplanar iff AB is perpendicular to

1 2b b

. That implies, 1 2AB. 0b b

2 1 1 2. 0a a b b

.

b) Cartesian form of co- planarity of lines:

We know that 2 1 2 1 2 1ˆˆ ÂB x i y y j z z k , 1 1 1 1

ˆˆ ˆb a i b j c k and

2 2 2 2ˆ

ˆ ˆb a i b j c k . So by using 2 1 1 2( ).( ) 0a a b b

, we get2 1 2 1 2 1

1 1 1

2 2 2

0

x x y y z z

a b c

a b c

.

Note that only coplanar lines can intersect each other in the plane they exist.

04. Angle between two planes: The angle between two planes is the angle between their normals 1m and 2m (say). Therefore if θ

is the angle between the planes 1 and 2 then o180 θ is also the angle between the two planes.

Though we shall be taking acute angle θ only as the angle between two planes.Observe the Fig.13.

a) Vector form for the angle between two planes:Consider the planes 1 1 1: .r m d

and 2 2 2: .r m d

. If θ is the angle between the normals

to the plane drawn from some common point. Then,

1 2

1 2

.cosθ

m mm m

[Using dot product of vectors

1 1 2

1 2

.θ cos

m m

m m

.

b) Cartesian form for the angle between two planes:Assume the planes, 1 1 1 1A B C D 0 x y z and 2 2 2 2A B C D 0 x y z where

1 1 1A , B , C and 2 2 2A , B , C are the d.r.’s of normals (to the planes) 1m and 2m respectively.

Then, 1 2 1 2 1 2

2 2 2 2 2 21 1 1 2 2 2

A A B B C Ccosθ

A B C A B C

+ +

+ + + +

1 1 2 1 2 1 2

2 2 2 2 2 21 1 1 2 2 2

A A B B C Cθ cos

A B C A B C

+ +

+ + + +

.

For the parallel planes, we have: 1 1 1

2 2 2

B C B C

.

For the perpendicular planes, we have: 1 2 1 2 1 2 0 A A B B C C .

05. Distance of a point from a plane: a) Vector form for the distance of a point from a plane:

Let : .r m d

be the plane and 1 1 1P , , y z be the point with position vector a . Let PA be

the length of perpendicular on the plane. See Fig.14. Since line PA passes through P a andis parallel to the m which is normal to the plane.So the vector equation of the line PA is ...( )r a m i





Since A is point of intersection of line (i) and the given plane. So we have,

.a m m d

2

.d a m

m

Putting the value of in (i), we get the position vector of A given as follow,

2

.d a mr a m

m

Since PA OA OP

PA r a

2

.PA

d a ma m a

m

2

.PA

d a mm

m

2

.PA PA

d a mm

m

2

.PA

d a mm

m

.. ., PA

d a mi e

m

Hence, length of the perpendicular PA ( ) say from a point having position vector a to the

plane .r m d is given by

.d a m p

m

.

b) Cartesian form for the distance of a point from a plane : Let A, B, C be the d.r.’s of the

normal m to the given plane. So by using the relation.d a m

pm

we can obtain,

1 1 1

2 2 2

A B C

A B C

x y z d p

.

If d is the distance from the origin and l, m, n are the d.c.’s of the normal vector to the plane through origin, then the coordinates of the foot of perpendicular is , ,ld md nd .

c) Distance between two parallel palnes: Assume the two planes as,

1 1. i.e., A B C D 0r m d x y z and 2 2. i.e., A B C D 0r m d x y z

.Then the distance p (say) between them is given as

(i) Vector form: 1 2d d p

m.

(ii) Cartesian form: 1 22 2 2

D D

A B C p .

06. Angle between a line and a plane:





The angle between a line and a plane is complementary to the angle between the line and normal to

the plane. Let θ is the angle between b (which is parallel to the line) and normal m of the plane.This implies that 90 θ is the angle between the line r a b

and plane .r m d

.Consider Fig.15.

Now the angle between b and m ,.

cosθb m

b m[By using dot product of vectors

So the angle ( ) say between the line and plane is given as 90 θ i.e.,

sin sin 90 θ cosθ

i.e.,.

sinb m

b m

1 .sin

b m

b m

.

This is the angle between line and a plane.

Hii, All!I hope this texture may have proved beneficial for you.While going through this material, if you noticed any error(s) or, something which doesn’t makesense to you, please bring it in my notice through SMS or Call at +91-9650 350 480 or Email [email protected] .

With lots of Love & Blessings!

- OP Gupta Electronics & Communications Engineering, Indira Award Winner www.theOPGupta.WordPress.com

Click on the following link to go for a pleasant surprise:http://theopgupta.wordpress.com/maths-rockers/



VARIOUS FIGURES RELATED TO THREE DIMENSIONAL GEOMETRY

Fig. 1 Fig. 2

Fig. 3 Fig. 4

Fig. 5 Fig. 6



Fig. 7 Fig. 8

Fig. 9 Fig. 10

Fig. 11 Fig. 12



Fig. 13

Fig. 14 Fig. 15

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New 3 Dimensional Geometry