Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 119 · 2015. 10. 25. · Prepared by: M. S....

Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 119 -


CHAPTER – 10: VECTOR ALGEBRA

MARKS WEIGHTAGE – 06 marks

QUICK REVISION (Important Concepts & Formulae) Vector

The line l to the line segment AB, then a magnitude is prescribed on the line l with one of the two directions, so that we obtain a directed line segment. Thus, a directed line segment has magnitude as well as direction.

A quantity that has magnitude as well as direction is called a vector. A directed line segment is a vector, denoted as AB

or simply as | |a

, and read as ‘vector AB

’ or

‘vector | |a

’. The point A from where the vector AB

starts is called its initial point, and the point B where it ends

is called its terminal point. The distance between initial and terminal points of a vector is called the magnitude (or length) of the vector, denoted as | |AB

or | |a

. The arrow indicates the direction of the

vector. The vector OP

having O and P as its initial and terminal points, respectively, is called the position

vector of the point P with respect to O. Using distance formula, the magnitude of vector OP

is given by 2 2 2| |OP x y z

The position vectors of points A, B, C, etc., with respect to the origin O are denoted by

,a b and c

, etc., respectively Direction Cosines

Consider the position vector ( )OP or r

of a point P(x, y, z) in below figure. The angles α, β, γ made by the vector r

with the positive directions of x, y and z-axes respectively, are called its direction

angles. The cosine values of these angles, i.e., cosα, cosβ and cos γ are called direction cosines of the vector r

, and usually denoted by l, m and n, respectively.


The triangle OAP is right angled, and in it, we have cos xr

(r stands for | r

|). Similarly, from the

right angled triangles OBP and OCP, we may write cos yr

and cos zr

. Thus, the coordinates

of the point P may also be expressed as (lr, mr, nr). The numbers lr, mr and nr, proportional to the direction cosines are called as direction ratios of vector r

, and denoted as a, b and c, respectively.

l2 + m2 + n2 = 1 but a2 + b2 + c2 ≠ 1, in general. Types of Vectors Zero Vector A vector whose initial and terminal points coincide, is called a zero vector (or null

vector), and denoted as 0

. Zero vector can not be assigned a definite direction as it has zero magnitude. Or, alternatively otherwise, it may be regarded as having any direction. The vectors

,AA BB

represent the zero vector, Unit Vector A vector whose magnitude is unity (i.e., 1 unit) is called a unit vector. The unit vector

in the direction of a given vector a

is denoted by a Coinitial Vectors Two or more vectors having the same initial point are called coinitial vectors. Collinear Vectors Two or more vectors are said to be collinear if they are parallel to the same line,

irrespective of their magnitudes and directions.

Equal Vectors Two vectors a and b

are said to be equal, if they have the same magnitude and direction regardless of the positions of their initial points, and written as a b

.


Negative of a Vector A vector whose magnitude is the same as that of a given vector (say, AB

), but direction is opposite to that of it, is called negative of the given vector. For example, vector BA

is negative of the vector AB

, and written as BA

= − AB

.

The vectors defined above are such that any of them may be subject to its parallel displacement

without changing its magnitude and direction. Such vectors are called free vectors. Addition of Vectors Triangle law of vector addition

If two vectors a and b

are represented (in magnitude and direction) by two sides of a triangle taken in order, then their sum (resultant) is represented by the third side c

( a b

) taken in the

opposite order.

Subtraction of Vectors : To subtract b

from a

, reverse the direction of b

and add to a

. Geometrical Representation of Addition and Subtraction :

Parallelogram law of vector addition

If we have two vectors a and b

represented by the two adjacent sides of a parallelogram in magnitude and direction, then their sum a b

is represented in magnitude and direction by the

diagonal of the parallelogram through their common point. This is known as the parallelogram law of vector addition.

Properties of vector addition Property 1

For any two vectors a and b

, a b b a

(Commutative property) Property 2

For any three vectors ,a b and c

, a b c a b c

(Associative property)

Property 3

For any vector a

, we have 0 0a a a

, Here, the zero vector 0

is called the additive identity for the vector addition.


Property 4 For any vector a

, we have 0a a a a

Here, the vector a

is called the additive inverse for the vector addition. Multiplication of a Vector by a Scalar

Let a

be a given vector and λ a scalar. Then the product of the vector a

by the scalar λ, denoted as λ a

, is called the multiplication of vector a

by the scalar λ. Note that, λ a

is also a vector, collinear to the vector a

. The vector λ a

has the direction same (or opposite) to that of vector a

according as

the value of λ is positive (or negative). Also, the magnitude of vector λ a

is |λ| times the magnitude of the vector a

, i.e., | λ a

| = | λ | | a

|

Unit vector in the direction of vector a

is given by 1 .| |

a aa

Properties of Multiplication of Vectors by a Scalar For vectors a

, b

and scalars m, n, we have (i) m ( a

) = (–m) a

= – (m a

)

(ii) (–m) ( a

) = m a

(iii) m(n a

) = (mn) a

= n(m a

)

(iv) (m + n) a

= m a

+ n a

(v) m( a

+ b

) = m a

+ mb

. Vector joining two points

If P1(x1, y1, z1) and P2(x2, y2, z2) are any two points, then the vector joining P1 and P2 is the vector

1 2PP

.

1 2PP

= Position vector of head – Position vector of tail = 2 1OP OP

= 2 1 2 1 2 1( ) ( ) ( )x x i y y j z z k

The magnitude of vector 1 2PP

is given by 2 2 22 1 2 1 2 1( ) ( ) ( )x x y y z z

SECTION FORMULA Internal Division : Position vector of a point C dividing a vector AB

internally in the ratio of m : n

is OC

= mb nam n

If C is the midpoint of AB

, then OC

divides AB

in the ratio 1 : 1. Therefore, position vector of C

is 1. 1.1 1 2b a a b


The position vector of the midpoint of AB

is 2

a b

Position vector of any point C on AB

can always be taken as c b a

where + = 1. . . ( ).n OA m OB n m OC

, where C is a point on AB

dividing it in the ratio m : n.

External Division : Let A and B be two points with position vectors a

and b

respectively and let C

be a point dividing AB

externally in the ratio m : n. Then the position vector of C is given by

OC

= mb nam n

Two vectors a

and b

are collinear if and only if there exists a nonzero scalar λ such that b

= λa. If

the vectors a

and b

are given in the component form, i.e. 1 2 3a a i a j a k

and

1 2 3b b i b j b k , then the two vectors are collinear if and only if

1 2 3 1 2 3b i b j b k a i a j a k

31 2

1 2 3

bb ba a a

If

1 2 3a a i a j a k , then a1, a2, a3 are also called direction ratios of a

.

In case if it is given that l, m, n are direction cosines of a vector, then li m j nk (cos ) (cos ) (cos )i j k is the unit vector in the direction of that vector, where α, β and γ are

the angles which the vector makes with x, y and z axes respectively. Product of Two Vectors Scalar (or dot) product of two vectors

The scalar product of two nonzero vectors a

and b

, denoted by .a b

, is defined as . | || | cosa b a b

where, θ is the angle between a

and b

, 0 ≤ θ ≤ π If either 0a

or 0b

, then θ is not defined, and in this case, we define . 0a b

Observations .a b

is a real number.

Let a

and b

be two nonzero vectors, then . 0a b

if and only if a

and b

are perpendicular to each other. i.e.

. 0a b a b

If θ = 0, then . | || |a b a b

. In particular, 2. | |a a a

, as θ in this case is 0.

If θ = π, then . | || |a b a b

In particular, 2.( ) | |a a a

, as θ in this case is π.

In view of the Observations 2 and 3, for mutually perpendicular unit vectors ,i j and k , we have . . . 1i i j j k k

. . . 0i j j k k i The angle between two nonzero vectors a

and b

, is given by

1. .cos , cos| || | | || |

a b a bora b a b

The scalar product is commutative. i.e. . .a b b a


Property 1 (Distributivity of scalar product over addition) Let ,a b and c

be any three vectors, then .( ) . .a b c a b a c

Property 2 Let a

and b

be any two vectors, and be any scalar. Then ( ). ( . ) .( )a b a b a b

Projection of a vector on a line If p is the unit vector along a line l, then the projection of a vector a

on the line l is given by .a p

.

Projection of a vector a

on other vector b

, is given by 1. . ( . )| | | |ba b or a or a bb b

If θ = 0, then the projection vector of AB

will be AB

itself and if θ = π, then the projection vector of AB

will be BA

If 2 or 3

2 , then the projection vector of AB

will be zero vector.

If α, β and γ are the direction angles of vector 1 2 3a a i a j a k

, then its direction cosines may be

given as 31 2.cos ,cos cos| | | | | || || |

aa aa ia a aa i

Vector (or cross) product of two vectors The vector product of two nonzero vectors a

and b

, is denoted by a

× b

and defined as | || | sina b a b n

, where, θ is the angle between a

and b

, 0 ≤ θ ≤ π and n is a unit vector

perpendicular to both a

and b

, such that a

, b

and n form a right handed system. If either 0a

or 0b

, then θ is not defined and in this case, we define 0a b

. Observations a

× b

is a vector. Let a

and b

be two nonzero vectors. Then 0a b

if and only if and a

and b

are parallel (or

collinear) to each other, i.e., 0 ||a b a b

In particular, 0a a

and ( ) 0a a

, since in the first situation, θ = 0 and in the second one, θ = π, making the value of sin θ to be 0.

If 2 then | || |a b a b

In view of the Observations 2 and 3, for mutually perpendicular unit vectors ,i j and k , we have 0i i j j k k

, ,i j k j k i k i j In terms of vector product, the angle between two vectors a

and b

may be given as

| |sin| || |a ba b

The vector product is not commutative, as a b b a

In view of the above observations, we have , ,j i k k j i i k j

If a

and b

represent the adjacent sides of a triangle then its area is given as 1 | |2

a b

.


If a

and b

represent the adjacent sides of a parallelogram, then its area is given by | |a b

.

The area of a parallelogram with diagonals a

and b

is 1 | |2

a b

| |a ba b

is a unit vector perpendicular to the plane of a

and b

.

| |a ba b

is also a unit vector perpendicular to the plane of a

and b

.

Property 3 (Distributivity of vector product over addition): If ,a b and c

are any three vectors

and λ be a scalar, then (i) ( )a b c a b a c

(ii) ( ) ( ) ( )a b a b a b

Let a

and b

be two vectors given in component form as

1 2 3a a i a j a k and

1 2 3b b i b j b k ,

respectively. Then their cross product may be given by

1 2 3

1 2 3

i j ka b a a a

b b b




NCERT Important Questions & Answers 1. Find the unit vector in the direction of the sum of the vectors, 2 2 5a i j k

and 2 3b i j k

Ans: The sum of the given vectors is ( , ) 4 3 2a b c say i j k

and 2 2 2| | 4 3 ( 2) 29c

Thus, the required unit vector is

1 1 4 3 2(4 3 2 )| | 29 29 29 29

c c i j k i j kc

2. Show that the points are (2 ), ( 3 5 ), (3 4 4 )A i j k B i j k C i j k the vertices of a right angled

triangle. Ans: We have (1 2) ( 3 1) ( 5 1) 2 6AB i j k i j k

(3 1) ( 4 3) ( 4 5) 2BC i j k i j k

and (2 3) ( 1 4) (1 4) 3 5CA i j k i j k

Then 2 2 2| | 41,| | 6,| | 35AB BC CA

2 2 2| | | | | |AB BC CA

Hence, the triangle is a right angled triangle. 3. Find the direction cosines of the vector joining the points A(1, 2, –3) and B(–1, –2, 1), directed

from A to B. Ans: The given points are A(1,2, − 3) and B(−1, −2,1). Then ( 1 1) ( 2 2) (1 ( 3)) 2 4 4AB i j k i j k

Now, | | 4 16 16 36 6AB

unit vector along AB

= 1 1 1 2 2( 2 4 4 )6 3 3 3| |

AB i j k i j kAB

Hence direction cosines are 1 2 2, ,3 3 3

4. Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are 2i j k and i j k respectively, in the ratio 2 : 1 (i) internally (ii) externally Ans: The position vector of a point R divided the line segment joining two points P and Q in the ratio m: n is given by

Case I Internally mb nam n

Case II Externally mb nam n

Position vectors of P and Q are given as 2 ,OP i j k OQ i j k


(i) Position vector of R [dividing (PQ ) in the ratio 2 : 1 internally]

2( ) 1( 2 ) 4 1 4 12 1 3 3 3 3

mOQ nOP i j k i j k i j k i j km n

(i) Position vector of R [dividing (PQ ) in the ratio 2 : 1 externally] 2( ) 1( 2 ) 3 0 3 3 3

2 1 1mOQ nOP i j k i j k i j k i k

m n

5. Find the position vector of the mid point of the vector joining the points P(2, 3, 4) and

Q(4, 1, –2). Ans: Position vectors of P and Q are given as 2 3 4 , 4 2OP i j k OQ i j k

The position vector of the mid point of the vector joining the points P(2, 3, 4) and Q(4, 1, –2) is given by

Position Vector of the mid-point of (PQ) = 1 1 4 2 2 3 42 2

OQ OP i j k i j k

1 6 4 2 3 22

i j k i j k

6. Show that the points A, B and C with position vectors, 3 4 4a i j k , 2b i j k

and 3 5c i j k

respectively form the vertices of a right angled triangle. Ans: Position vectors of points A, B and C are respectively given as

3 4 4 , 2 3 5a i j k b i j k and c i j k

Now, 2 3 4 4 3 5AB b a i j k i j k i j k

2| | 1 9 25 35AB

3 5 2 2 6BC c b i j k i j k i j k

2| | 1 4 36 41BC

3 4 4 3 5 2CA a c i j k i j k i j k

2| | 4 1 1 6CA

2 2 2| | | | | |BC AB CA

Hence it form the vertices of a right angled triangle. 7. Find angle ‘θ’ between the vectors a i j k

and b i j k

Ans: The angle θ between two vectors a

and b

is given by

.cos| || |

a ba b

Now, . ( ).( ) 1 1 1 1a b i j k i j k

Therefore, we have 11 1cos cos3 3

8. If 5 3a i j k and 3 5b i j k

, then show that the vectors a b

and a b

are perpendicular. Ans: We know that two nonzero vectors are perpendicular if their scalar product is zero. Here, 5 3 3 5 6 2 8a b i j k i j k i j k

and 5 3 3 5 4 4 2a b i j k i j k i j k


Now, ( ).( ) (6 2 8 ).(4 4 2 ) 24 8 16 0a b a b i j k i j k

Hence a b

and a b

are perpendicular. 9. Find | |a b

, if two vectors a

and b

are such that | | 2a

, | | 3b

and .a b

= 4. Ans: We have

2| | ( ).( ) . . . .a b a b a b a a a b b a b b

2 2 2 2| | 2( . ) | | 2 2(4) 3 4 8 9 5a a b b

| | 5a b

10. Show that the points ( 2 3 5 ), ( 2 3 ), (7 )A i j k B i j k C i k are collinear. Ans: We have

(1 2) (2 3) (3 5) 3 2AB i j k i j k

(7 1) (0 2) ( 1 3) 6 2 4BC i j k i j k

(7 2) (0 3) ( 1 5) 9 3 6CA i j k i j k Now, 2 2 2| | 14,| | 56,| | 126AB BC CA

| | 14,| | 2 14,| | 3 14AB BC CA

| | | | | |CA AB BC

Hence the points A, B and C are collinear.

11. If , ,a b c

are unit vectors such that 0a b c

, find the value of . . .a b b c c a

Ans: Given that | | 1,| | 1,| | 1, 0a b c a b c

2( ) ( ).( ) 0a b c a b c a b c

. . . . . . . . . 0a a a b a c b b b c b a c a c b c c

2 2 2| | | | | | 2( . . . ) 0a b c a b b c c a

1 1 1 2( . . . ) 0a b b c c a

2( . . . ) 3a b b c c a

3. . .

2a b b c c a

12. If the vertices A, B, C of a triangle ABC are (1, 2, 3), (–1, 0, 0), (0, 1, 2), respectively, then find

ABC. Ans: We are given the points A(1, 2, 3), B(−1, 0, 0) and C(0, 1, 2). Also, it is given that ABC is the angle between the vectors BA

and BC

Now, ( 2 3 ) ( 0 0 ) 2 2 3BA i j k i j k i j k

| | 4 4 9 17BA

and (0 2 ) ( 0 0 ) 2BC i j k i j k i j k

| | 1 1 4 6BC

. (2 2 3 ).( 2 ) 2 2 6 10BA BC i j k i j k

.cos

| || |BA BC

BA BC

10 10cos

( 17)( 6) 102ABC


1 10cos102

ABC

13. Show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, –1) are collinear.

Ans: The given points are A(1, 2, 7), B (2, 6, 3) and C(3, 10, −1).

(2 6 3 ) ( 2 7 ) 4 4AB i j k i j k i j k

| | 1 16 16 33AB

(3 10 ) (2 6 3 ) 4 4BC i j k i j k i j k

| | 1 16 16 33BC

and (3 10 ) ( 2 7 ) 2 8 8AC i j k i j k i j k

| | 4 64 64 132 2 33AC

| | | | | |AC AB BC

Hence, the given points A, B and C are collinear.

14. Show that the vectors 2i j k , 3 5i j k and 3 4 4i j k form the vertices of a right angled triangle. Ans: Let A= 2i j k B = 3 5i j k and C = 3 4 4i j k

( 3 5 ) (2 ) 2 6AB i j k i j k i j k

| | 1 4 36 41AB

(3 4 4 ) ( 3 5 ) 2BC i j k i j k i j k

| | 4 1 1 6BC

and (3 4 4 ) (2 ) 3 5AC i j k i j k i j k

| | 1 9 25 35AC

2 2 2| | | | | |AB AC BC

Hence, ABC is a right angled triangle.

15. Find a unit vector perpendicular to each of the vectors ( )a b

and ( )a b

, where , 2 3a i j k b i j k

. Ans: We have 2 3 4a b i j k

and 2a b j k

A vector which is perpendicular to both ( )a b

and ( )a b

is given by

( ) ( ) 2 3 4 2 4 2 ( , )

0 1 2

i j ka b a b i j k c say

Now, | | 4 16 4 24 2 6c

Therefore, the required unit vector is

1 1 1 2 2( 2 4 2 )| | 2 6 6 6 6

c c i j k i j kc

16. Find the area of a triangle having the points A(1, 1, 1), B(1, 2, 3) and C(2, 3, 1) as its vertices. Ans: We have 2AB j k

and 2AC i j

.


The area of the given triangle is 1 | |2

AB AC

Now,

0 1 2 4 2

1 2 0

i j kAB AC i j k

Therefore, | | 16 4 1 21AB AC

Thus, the required area is 1 1| | 212 2

AB AC

17. Find the area of a parallelogram whose adjacent sides are given by the vectors 3 4a i j k

and b i j k .

Ans: The area of a parallelogram with a

and b

as its adjacent sides is given by | |a b

Now,

3 1 4 5 4

1 1 1

i j ka b i j k

Therefore, | | 25 1 16 42a b

and hence, the required area is 42 .

18. Find the area of the triangle with vertices A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5).

Ans: (2 3 5 ) ( 2 ) 2 3AB i j k i j k i j k

( 5 5 ) ( 2 ) 4 3AC i j k i j k j k

Now,

1 2 3 6 3 4

0 4 3

i j kAB AC i j k

| | 36 9 16 61AB AC

Area of triangle ABC = 1 61| |2 2

AB AC

sq. units.

19. Find the area of the parallelogram whose adjacent sides are determined by the vectors 3a i j k

and 2 7b i j k .

Ans: Adjacent sides of parallelogram are given by the vectors 3a i j k

and 2 7b i j k .

Now,

1 1 3 20 5 5

2 7 1

i j ka b i j k

| | 400 25 25 450 15 2a b

Hence, the area of the given parallelogram is 15 2 sq. units.

20. Let the vectors a

and b

be such that | | 3a

and 2| |3

b

, then a b

is a unit vector, find the

angle between a

and b

. Ans:


Given that vectors a

and b

be such that | | 3a

and 2| |3

b

.

Also, a b

is a unit vector | | 1a b

2| | . | | sin 1 3 sin 1

3a b

1sin42

21. If i j k , 2 5i j , 3 2 3i j k and 6i j k are the position vectors of points A, B, C and D respectively, then find the angle between AB and CD . Deduce that AB and CD are collinear. Ans: Note that if θ is the angle between AB and CD , then θ is also the angle between AB and CD . Now AB = Position vector of B – Position vector of A = (2 5 ) ( ) 4i j i j k i j k Therefore, | | 1 16 1 18 3 2AB

Similarly, 2 8 2 | | 4 64 4 72 6 2CD i j k CD

Thus, . 1( 2) 4( 8) ( 1)(2)cos 1| || | (3 2)(6 2)

AB CDAB CD

Since 0 ≤ θ ≤ π, it follows that θ = π. This shows that AB and CD are collinear. 22. Let a

, b

and c

be three vectors such that | | 3,| | 4,| | 5a b c

and each one of them being perpendicular to the sum of the other two, find | |a b c

.

Ans: Given that each one of them being perpendicular to the sum of the other two. Therefore, .( ) 0, .( ) 0, .( ) 0a b c b c a c a b

Now, 2 2| | ( ) ( ).( )a b c a b c a b c a b c

. .( ) . .( ) .( ) .a a a b c b b b c a c a b c c

2 2 2| | | | | |a b c

9 16 25 50 Therefore, | | 50 5 2a b c

23. Find a vector of magnitude 5 units, and parallel to the resultant of the vectors 2 3a i j k

and 2b i j k

. Ans: Given vectors 2 3a i j k

and 2b i j k .

Let c

be the resultant vector a

and b

then (2 3 ) ( 2 ) 3 0c i j k i j k i j k

| | 9 1 0 10c

Unit vector in the direction of c

= 1 1 (3 )| | 10

c c i jc

Hence, the vector of magnitude 5 units and parallel to the resultant of vectors a

and b

is

1 3 10 105 5 (3 )2 210

c i j i j


24. The two adjacent sides of a parallelogram are 2 4 5i j k and 2 3i j k . Find the unit vector parallel to its diagonal. Also, find its area. Ans: Two adjacent sides of a parallelogram are given by 2 4 5a i j k

and 2 3b i j k

Then the diagonal of a parallelogram is given by c a b

2 4 5 2 3 3 6 2c a b i j k i j k i j k | | 9 36 4 49 7c

Unit vector parallel to its diagonal = 1 1 3 6 2(3 6 2 )7 7 7 7| |

c c i j k i j kc

Now,

2 4 5 22 11 0

1 2 3

i j ka b i j k

Then the area of a parallelogram = | | 484 121 0 605 11 5a b

sq. units.

25. Let 4 2 , 3 2 7a i j k b i j k and 2 4c i j k

. Find a vector d

which is perpendicular to both a

and b

and . 15c d

.

Ans: The vector which is perpendicular to both a

and b

must be parallel to a b

.

Now,

1 4 2 32 14

3 2 7

i j ka b i j k

Let ( ) (32 14 )d a b i j k

Also . 15 (2 4 ). (32 14 ) 15c d i j k i j k

15 564 56 15 9 159 3

Required vector 5 (32 14 )3

d i j k

26. The scalar product of the vector i j k with a unit vector along the sum of vectors 2 4 5i j k and 2 3i j k is equal to one. Find the value of λ.

Ans: Let a

= i j k , b

= 2 4 5i j k and c

= 2 3i j k

Now, 2 4 5 2 3 (2 ) 6 2b c i j k i j k i j k

2 2 2| | (2 ) 36 4 4 4 40 4 44b c

Unit vector along b c

is

2

(2 ) 6 2| | 4 44b c i j kb c

The scalar product of i j k with this unit vector is 1.

2

(2 ) 6 2( ). 1 ( ). 1| | 4 44b c i j ki j k i j kb c

2 2

(2 ) 6 2 61 14 44 4 44

2 2 26 4 44 ( 6) 4 44 2 212 36 4 44 8 8 1


27. If a

, b

and c

are mutually perpendicular vectors of equal magnitudes, show that the vector a b c

is equally inclined to a

, b

and c

. Ans: Given that a

, b

and c

are mutually perpendicular vectors. . . . 0a b b c c a

It is also given that | | | | | |a b c

Let vector a b c

be inclined to a

, b

and c

at angles andrespectively. 2( ). . . . | | 0 0cos

| || | | || | | || |a b c a a a b a c a a

a b c a a b c a a b c a

2| | | || || | | |

a aa b c a a b c

2( ). . . . 0 | | 0cos| || | | || | | || |

a b c b a b b b c b ba b c b a b c a a b c a

2| | | || || | | |

b ba b c a a b c

2( ). . . . 0 0 | |cos| || | | || | | || |

a b c c a c b c c c ca b c c a b c a a b c a

2| | | || || | | |

c ca b c a a b c

Now as | | | | | |a b c

, therefore, cos = cos = cos Hence, the vector a b c

is equally inclined to a

, b

and c

. .




Previous Years Board Exam (Important Questions & Answers) 1. Write the projection of vector i j k along the vector j .

Ans:

Required projection

( ). 0 1 0 1 110 1 0| |

i j k jj

2. Find a vector in the direction of vector 2 3 6i j k which has magnitude 21 units.

Ans:

Required vector 2 3 6 2 3 621 21

4 9 36 49i j k i j k

2 3 621 3(2 3 6 ) 6 9 187

i j k i j k i j k

3. Show that the vectors a b

, b c

and c a

are coplanar if , ,a b c

are coplanar. Ans: Let , ,a b c

are coplanar then we have 0a b c

.( ) .( ) .( ) 0a b c b c a c a b

Now, ( ).{( ) ( )}a b b c c a a b b c c a

( ).{ }a b b c b a c c c a

( ).{ }a b b c b a c a

.( ) .( ) .( ) .( ) .( ) .( )a b c a b a a c a b b c b b a b c a

0 0 0 0a b c b c a

a b c a b c

2 2 0 0a b c

Hence, , ,a b c

are coplanar 4. Show that the vectors , ,a b c

are coplanar if a b

, b c

and c a

are coplanar. Ans: Let , ,a b b c c a

are coplanar

( ).{( ) ( )} 0a b b c c a

( ).{ } 0a b b c b a c c c a

( ).{ } 0a b b c b a c a

.( ) .( ) .( ) .( ) .( ) .( ) 0a b c a b a a c a b b c b b a b c a

0 0 0 0 0a b c

0a b c

, ,a b c

are coplanar


5. Write a unit vector in the direction of vector PQ

, where P and Q are the points (1, 3, 0) and (4, 5, 6) respectively. Ans:

(4 1) (5 3) (6 0) 3 2 6PQ i j k i j k

Required unit vector = 3 2 6 3 2 6 3 2 6 3 2 6

7 7 7 79 4 36 49i j k i j k i j k i j k

6. Write the value of the following : ( ) ( ) ( )i j k j k i k i j Ans: ( ) ( ) ( )i j k j k i k i j i j i k j k j i k i k j

0k j i k j i 7. Find the value of 'p' for which the vectors 3 2 9i j k and 2 3i p j k are parallel.

Ans: Since given two vectors are parallel.

3 2 9 3 21 2 3 1 2p p

16 23

p p

8. Find .( )a b c

, if 2 3 , 2a i j k b i j k

and 3 2c i j k .

Ans: Given that 2 3 , 2a i j k b i j k

and 3 2c i j k

2 1 3.( ) 1 2 1 2(4 1) 1( 2 3) 3( 1 6)

3 1 2a b c

6 5 21 10 9. Show that the four points A, B, C and D with position vectors 4 5 , ,3 9 4i j k j k i j k and

4( )i j k are coplanar. Ans: Position vectors of A, B, C and D are

Position vector of A = 4 5i j k

Position vector of B = j k

Position vector of C = 3 9 4i j k

Position vector of D = 4( )i j k = 4 4 4i j k 4 6 2 , 4 3 , 8 3AB i j k AC i j k AD i j k

Now, 4 6 2

.( ) 1 4 3 4(12 3) 6( 3 24) 2(1 32)8 1 3

AB AC AD

60 126 66 0 .( ) 0AB AC AD

Hence ,AB AC and AD

are coplanar i.e. A, B, C and D are coplanar.


10. Find a vector a

of magnitude5 2 , making an angle of 4 with x-axis.,

2 with y-axis and an

acute angle with z-axis. Ans: Direction cosines of required vector a

are

1cos4 2

l , cos 0

2m and cosn

2 2 2 1l m n 2

2 21 1 10 cos 1 cos 12 22

1 1cos2 2

n

Unit vector in the direction of 1 102 2

a i j k

1 15 2 0 5 52 2

a i j k i k

11. If a

and b

are perpendicular vectors, | a b

| = 13 and | a

| = 5 find the value of |b

|. Ans: Given | a b

| = 13

2| | 169 ( ).( ) 169a b a b a b

2 2| | 2 . | | 169a a b b

2 2| | | | 169 . 0a b a b a b

2 2| | 169 | | 169 25 144b a

| | 12b

12. Find the projection of the vector 3 7i j k on the vector 2 3 6i j k .

Ans: Let a

= 3 7i j k and b

= 2 3 6i j k

Projection of the vector a

on b

=

. ( 3 7 ).(2 3 6 )

| | | 2 3 6 |a b i j k i j kb i j k

2 9 42 35 35 574 9 36 49

13. If a

and b

are two unit vectors such that a b

is also a unit vector, then find the angle

between a

and b

. Ans: Given that a b

is also a unit vector

| a b

| = 1 2| | ( ).( )a b a b a b

2 2 2| | 2 . | | 1 1a a b b

1 2 . 1 1 | | 1,| | 1a b a b

1 12 . 1 . | || | cos2 2

a b a b a b


1 1 21 1 cos cos cos cos2 2 3

23

14. Prove that, for any three vectors , ,a b c

2a b b c c a a b c

Ans: ( ).{( ) ( )}a b b c c a a b b c c a

( ).{ }a b b c b a c c c a

( ).{ }a b b c b a c a

.( ) .( ) .( ) .( ) .( ) .( )a b c a b a a c a b b c b b a b c a

0 0 0 0a b c b c a

a b c a b c

2 a b c

15. Vectors , ,a b c

are such that a b c

= 0 and | | 3, | | 5 | | 7a b and c

. Find the angle between a

and b

.

Ans: 0a b c a b c

2 2( ) ( )a b c

| ( ).( ) .a b a b c c

2 2 2| | 2 . | | | | 9 2 . 25 49a a b b c a b

2 . 49 25 9 15a b

15 15. | || | cos2 2

a b a b

15 13 5 cos cos cos cos2 2 3

3

16. If a

is a unit vector and ( )( ) 24x a x a

, then write the value of | x

|.

Ans: Given that ( )( ) 24x a x a

. . . . 24x x x a a x a a

2 2| | | | 24 . .x a x a a x

2 2| | 1 24 | | 25 | | 5x x x

17. For any three vectors ,a b and c

, write the value of the following:

( ) ( ) ( )a b c b c a c a b

Ans:

( ) ( ) ( )a b c b c a c a b

a b a c b c b a c a c b

0a b a c b c a b a c b c


18. The magnitude of the vector product of the vector i j k with a unit vector along the sum of vectors 2 4 5i j k and 2 3i j k is equal to 2 . Find the value of . Ans: Let a

= i j k , b

= 2 4 5i j k and c

= 2 3i j k

Now, 2 4 5 2 3 (2 ) 6 2b c i j k i j k i j k

2 2 2| | (2 ) 36 4 4 4 40 4 44b c

The vector product of i j k with this unit vector is 2 .

( )2 2| | | |b c a b cab c b c

Now,

( ) 1 1 1 ( 2 6) ( 2 2 ) (6 2 )

2 6 2

i j ka b c i j k

8 (4 ) (4 )i j k

2

2 2

2

( ) 8 (4 ) (4 )2 2| | 4 44

64 (4 ) (4 )2

4 44

a b c i j kb c

2 2 2 2 2

2 2 2

64 (4 ) (4 ) 64 16 8 16 8 96 22 2 24 44 4 44 4 44

2 2 2 296 2 2( 4 44) 96 2 2 8 88

8 8 1 19. Find a unit vector perpendicular to each of the vectors 2a b

and 2a b

, where

3 2 2a i j k and 2 2b i j k

. Ans. Given that 3 2 2a i j k

and 2 2b i j k

2 3 2 2 2( 2 2 )a b i j k i j k

3 2 2 2 4 4 5 6 2i j k i j k i j k

and 2 2(3 2 2 ) 2 2a b i j k i j k 6 4 4 2 2 7 6 2i j k i j k i j k

Now, perpendicular vector of 2a b

and 2a b

=

5 6 2 (12 12) (10 14) (30 42)

7 6 2

i j ki j k

24 24 12 12(2 2 )i j k i j k

Required unit vector = 12(2 2 ) 2 2

12 4 4 1 9i j k i j k

2 2 2 2 13 3 3 3

i j k i j k


20. If 7a i j k and 5b i j k

, then find the value of , so that a b

and a b

are perpendicular vectors. Ans: Given that 7a i j k

and 5b i j k

7 5 6 2 (7 )a b i j k i j k i j k

and 7 5 4 (7 )a b i j k i j k i k

Now, a b

and a b

are perpendicular vectors ( ).( ) 0a b a b

(6 2 (7 ) ).( 4 (7 ) 0i j k i k

24 0 (7 )(7 ) 0 2 224 49 0 25 5

Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 119 · 2015. 10. 25. · Prepared by: M. S....

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Transcript of Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 119 · 2015. 10. 25. · Prepared by: M. S....