12_Vectors.pdf

VECTORS

1

VECTORS 1.1

Basic concept · Vectors have magnitude and direction, whereas scalars have only magnitude. ·

®

Vector. A directed line segment represents a vector. A vector from P to Q is denoted by PQ . P ®

and Q are called respectively initial and terminal points of the vector PQ . The direction of vector ®

PQ

· ·

is from P to Q. ®

®

The magnitude of vector PQ is denoted by | PQ | and represents length of line segment PQ. Zero vector. A vector whose initial and terminal points coincide is called a zero vector. It is also ®

r

®

r

r

defined as a vector whose magnitude is zero. It is denoted by 0 or AA, BB etc. | 0 | = 0, | 0 | = 0, ®

·

| AA | = 0. r Unit vector. A vector a is called a unit vector if its magnitude is one unit. It is denoted by aˆ , | aˆ | = 1. r a r r Unit vector represents direction along a vector a , also aˆ = a .

·

Position vector of a point. If we take a fixed point O, which is called the origin of reference ®

and P be any point in the plane, then the vector OP is called the position vector of P relative to r O. If position vector of point P is a , denoted as P( ar ), then corresponding to some origin of ®

®

reference O, OP = a . ®

1.2

· Line of support. A line, whose segment is PQ, is called the line of support of the vector PQ . Types of vectors 1. Like vectors. Two vectors are said to be like vectors, if they have (i) same or parallel lines of support. (ii) same direction. 2. Unlike vectors. Two vectors are said to be unlike vectors, if (i) their lines of support are same or parallel, (ii) opposite direction. 3. Equal vectors. Two vectors are said to be equal, if they have (i) same or parallel lines of support, (ii) same direction and (iii) equal magnitudes. We can also say that if like vectors have equal magnitudes then they are equal. 4. Coinitial vectors. Two or more vectors are said to be coinitial vectors if they have the same ®

5. 6. 7.

®

®

initial point, e.g., AP, AQ, AR etc. are coinitial vectors. Collinear vectors. Two or more vectors are called collinear vectors, if they have the same or parallel lines of support. Coplanar vectors. Any number of non-zero vectors are said to be coplanar if they lie in the same plane or parallel planes. Negative vector. A vector which has the same magnitude but opposite direction of the given ®

1.3

®

®

®

vector, is called the negative of the given vector. If PR = x then RP = - x . Properties with respect to addition · Triangle law of vector addition. If two vectors are represented along two sides of a triangle taken in order, then their resultant is represented by the third side taken in the opposite order. ®

If the sides OA and AB of DOAB represent OA and , ®

®

®

®

®

®

then OA + AB = OB , i.e., OB = a + b

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O

B r b

r a

A

VECTORS

2 ·

Parallelogram law of vector addition. If two vectors are represented along two adjacent sides of a parallelogram, then their resultant is represented along the diagonal of a parallelogram passing through the common vertex of adjacent sides. If the sides OA and OC of a parallelogram OABC represent respectively ®

OA

·

®

®

®

O

A

®

and OC , then OA + OC = OB .

r

r

r

Additive identity. For every vector a , a zero vector 0 is its additive identity as ar + 0 = ar . r Additive inverse. For a vector ar , a negative of vector ar is its additive inverse as ar + (– ar ) = 0 . ®

1.5

B

Properties with respect to addition r r r (i) Commutative. For vectors ar and b , we have ar + b = b + ar . r r r (ii) Associative. For vectors ar , b and cr , we have ar + ( b + cr ) = ( ar + b ) + cr (iii) (iv)

1.4

C

®

®

®

®

r

®

®

Also for vector AB , its additive inverse is BA as AB + BA = AA = 0 , Also, we conclude BA = – AB . Properties of with respect to multiplication · Multiplication of a vector by a scalar. Let ar be a given vector and k a scalar, then multiplication of vector ar by scalar k, denoted k ar , is a vector whose magnitude is k times that of vector ar and direction is (i) same as that of ar , if k > 0. (ii) opposite to that of ar , if k < 0. (iii) a zero vector, if k = 0. r r · To prove ar is parallel to b we have to show that ar = k b , where k is a scalar.. · Properties with respect to multiplication of a vector by a scalar r For vectors ar , b and scalars l, m, we have r r (i) l( ar + b ) = l ar + l b (ii) (l + m) ar = l ar + m ar (iii) l(m ar ) = (lm) ar . Propertes of vectors · To find a vector when position vectors of end points are given : r Let ar and b be the position vectors of end points A and B of a line segment AB. Then, ®

®

· ·

®

r

= Position vector of B – Position vector of A = OB – OA = b – ar . r r Components of a vector. If rr = ar + b , then ar and b are known as components of rr . Vector in two dimensions : Let rr be position vector of a point P(x, y), AB

(i)

® Then rr = OP = x ˆi + y ˆj

(ii)

x ˆi and y ˆj are known as component vectors of rr along x and y-axis and x and y are known as components of rr along x and y-axis.

(iii)

Magnitude of rr , is given by, | rr | = x 2 + y 2 .

(iv)

r x y x ˆi + yˆj r Unit vector along rr , is given by, rr = r = rr , i.e., rˆ = rr ˆi + rr ˆj . r

(v)

y x If q is angle which rr makes with x-axis, then cos q = r , sin q = r . r

(vi) (vii)

r

r r.

cos q, sin q are known as direction cosines of For rr = x ˆi + y ˆj ; x and y are known as direction ratios of rr , i.e., components of a vector are direction ratios of rr , but converse may or may not be true.

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3

·

VECTORS Vector in three dimensions : r If r is position vector of point P(x, y, z) (i)

r

Then r = xˆi + yˆj + zkˆ, xˆi, yˆj and zkˆ are component vectors and x, y Z kˆ g

O ˆi

a

r r b

Y

ˆj

X

and z are components of vector rr along x, y and z-axis. (ii) (iii)

Magnitude of rr , is given by, | rr | = x 2 + y 2 + z 2 If a, b, g are the angles which vector rr makes with x, y and z-axis, then its direction x

y

z

consies cos a, cos b, cos g are cos a = | rr | ; cos b = | rr | ; cos g = | rr | . (iv)

x y z Unit vector along rr , is given by, rr = | rr | ˆi + | rr | ˆj + | rr | kˆ = (cos a) ˆi + (cos b) ˆj +

(cos g) kˆ

·

r

If r = xˆi + yˆj + zkˆ , then x, y, z are direction ratios of rr , i.e., components of a vector are direction ratios of a vector. Vector when coordinates of end points are given. If A (x1, y1, z1) and B(x2, y2, z2) be the

(v)

®

end-points of line segment AB, then AB = (x2 – x1) ˆi + (y2 – y1) ˆj + (z2 – z1) kˆ .

·

® (x2 – x1), (y2 – y1) and (z2 – z1) are components of AB along x, y and z-axis respectively. These are also direction ratios along x, y and z-axis respectively. Position vector of a point dividing the line segment in a given ratio (Section formula). Position vector rr of a point which divides the join of two given points with position vectors ar

r

r

r l b + ma and b in the ratio l : m internally is, rr = . l+m

If point divides externally, in the ratio l : m, then position vector is ·

1.6

r r l b - ma . l -m

Position vector of the mid-point (1 : 1) of the line segment joining the end points with position r r r a+b r zvectors a and b is . 2

Scalar or dot product of two vectors r · Scalar or dot product of two vectors. If q is the angle between two given vectors ar and b , then r

· · · · · · · ·

r

r

their scalar or dot product, denoted by ar . b is given by ar . b = | ar | | b | cos q, where q is r angle between ar and b . r r a . b is a scalar quantity.. r ( ar . b ). cr is not defined. r r r Commutative property. For vectors ar and b , ar . b = b . ar . r r Associative property. For vectors ar , b , cr , associative property does not hold as ar . ( b . cr ) is not defined. r r r Distributive property. For vectors ar , b , cr , ; ar .( b + cr ) = ar . b + ar . cr . r r r r For vectors ar and b , ar .(l b ) = (l ar ) . b = l( ar . b ), l is a scalar.. r r r r r r For vectors ar and b , ar . b = 0 Û ar = 0 , b = 0 or ar ^ b . For vector ar , ar . ar = ar 2 = | ar |2, i.e., square of a vector is equal to square of its magnitude.

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VECTORS

4 r

r and projection vector of ar along b is

r r æ a .b ö r ç r ÷ ç | b |2 ÷ b . è ø

·

Projection of ar along b is

·

r r a.b r r r Angle between a and b is given by, cos q = r r , q is the angle between ar and b . | a ||b|

·

If ˆi , ˆj , kˆ are unit vectors along x, y and z-axis respectively, then ˆi . ˆi = ˆj . ˆj = kˆ . kˆ = 1 ; ˆi . ˆj = 0, ˆj . kˆ = 0, kˆ . ˆi = 0.

·

r r r If ar = a1 ˆi + b1 ˆj + c1 kˆ and b = a2 ˆi + b2 ˆj + c2 kˆ , then ar . b = a1a2 + b1b2 + c1c2 If ar ^ b then

·

r r a . b = 0 Þ a1a2 + b1b2 + c1c2 = 0 r If q is angle between ar = a1 ˆi + b1 ˆj + c1 kˆ and b = a2 ˆi + b2 ˆj + c2 kˆ , then a1a2 + b1b 2 + c1c 2

cos q = 1.7

r r a .b r |b|

a12

+ b12 + c12 a22 + b22 + c 22

.

Vector or cross product of two vectors r r · Vector or cross product of two vectors ar and b . denoted by ar . b , is given by r r r r r r r r a × b = | a | | b | sin q n, q is angle between a and b , nˆ is a unit vector ^ to a × b , is given by r r r r a and b and direction is such that a , b and nˆ form a right hand system. r r r r r r · a × b is a vector quantity, whose magnitude is, | a × b | = | a | | b | sin q. · · · · · · · ·

r b.

r r | a´b | r r If q is angle between a and b , then sin q = r r . | a || b | r r r r For a , a × a = 0 . r r r For vectors ar and b , ar × b = – b × ar , i.e., cross product of two vectors is not commutative. r r r r r r r For ar , b and cr , a × ( b × c ) ¹ ( a × b ) × c , in general. Not associative. r r r Distributive property. For vectors ar , b and cr , ar × ( b + cr ) = ar × b + ar × cr . r r r r r r r For vectors ar and b , ar × b = 0 Û ar = 0 , b = 0 or ar || b . r If we want to show that two non-zero vectors ar and b are parallel, then we should r r show that ar × b = 0 . r Geometrically, | ar × b | represents area of a parallelogram whose adjacent sides are along ar and

·

r r 1 Area of a triangle whose sides are along ar and b is given by | ar × b |.

·

Area of a parallelogram whose diagonals are along d1 and d2 is given by

·

If ˆi , ˆj , kˆ are vectors along x, y and z-axis respectively, then

·

× ˆi = 0 , ˆj × ˆj = 0 , kˆ × kˆ = 0 ˆi × ˆj = kˆ ; ˆj × kˆ = ˆi ; kˆ + ˆi = ˆj . r r r r For a scalar l, l( ar × b ) = (l ar × b ) = ( ar × l b ), where ar and b are given vectors.

2

r

ˆi

r

r

r

r

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1 r |d 2 1

r

× d2 |.

VECTORS

5

SOLVED PROBLES Ex.1

r Find the projection of ar = iˆ – 3 kˆ on b = 3 iˆ + ˆj - 4 kˆ .

Sol.

r r 15 r ar.b 3 + 0 + 12 r Projection of a on b = = . = 9 + 1 + 16 26 |b|

Ex.2

If ar is a unit vector and ( xr + ar ).( xr – ar ) = 15, find | xr |.

Sol.

x 2 - a2 = 15

Ex.3

Find a vector in the direction of vector 5 ˆi - ˆj - 2ˆ k which has magnitude 8 units.

Sol.

The unit vector in the direction of vector a = 5ˆi - ˆj + 2kˆ is

®

®

Þ| x | =15+| ar | =16 Þ| x |= 4 r 2

2

r

r

1 r |a|

1

. ar =

( 5ˆi - ˆj + 2kˆ )

2

5 + ( -1)2 + 22

1

1

( 5ˆi - ˆj + 2kˆ ) = 30 ( 5ˆi - ˆj + 2kˆ ) 25 + 1 + 4

=

8 The vector of 8 units in the direction of the vector ar =

30

Ex.4

( 5ˆi - ˆj + 2kˆ )

Find a vector of magnitude 5 units, and parallel to the resultant of the vectors ar = 2 ˆi + 3 ˆj - kˆ r

and b = iˆ - 2 ˆj + kˆ . Sol.

r

r

The resultant of vector a = 2ˆi + 3ˆj - kˆ and b = ˆi - 2ˆj + kˆ is given by r r a+b

= (2ˆi + 3ˆj - kˆ ) + (ˆi - 2ˆj + kˆ ) = 3ˆi + ˆj r

The unit vector in the direction of ar + b is =

3 2

2

3 +1

1

ˆi +

3

2

3 + 12

ˆj

i.e.,

1

= 10 ˆi + 10 ˆj r

A vector of magnitude 5 units, and parallel to ar + b is, therefore, i.e., Ex.5

IF

=

3 2

r r a = iˆ + ˆj + kˆ , b = 2 iˆ - ˆj + 3 kˆ

r

r

r

Here, 2a = 2ˆi + 2ˆj + 2kˆ - b = -2ˆi + ˆj + 3kˆ and

r r r 2a - b + 3c = 3ˆi - 3ˆj + 2kˆ

r 3c = 3ˆi - 6ˆj + 3kˆ

3 ˆ 3 ˆ 2 ˆ r ij+ k The unit vector parallel to the vector 2ar - b + 3cr = 22 22 22

Ex.6

Show that the vectors 2 iˆ - 3 ˆj + 4 kˆ and -4 iˆ - 6 ˆj - 8 kˆ are collinear..

Sol.

Two vectors a = a1ˆi + a 2 ˆj + a3kˆ and

r

r b = b1ˆi - b2 ˆj + b3kˆ

ö

AND cr = iˆ - 2 ˆj + kˆ , FIND A UNIT VECTOR PARALLEL TO THE

r

VECTOR 2 a - b + 3c . Sol.

1

10 ˆ j 2

10 ˆi +

r

æ 3

= 5 çç 10 ˆi + 10 ˆj ÷÷ ø è

a

a

b

3 1 2 are collinear if b = b = b 1 2 3

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VECTORS

6 For given vectors, we have Each is equal to

2 -4

=

-3 6

=

4 -8

-1 . Hence , the given vectors are collinear.. 2

Ex.7

Find the direction cosines of the vector ˆi + 2 ˆj + 3 kˆ .

Sol.

For a vector a1ˆi + a2 ˆj + a3kˆ , a1, a2, a3 are the direction ratios of the vector.. Hence, for the given vector, a1 = 1, a 2 = 2 and a3 = 3. 1 2

2

1 +2 +3

2

,

2 2

2

1 +2 +3

2

,

3 2

1 + 22 + 3 2

are the required direction cosines. 1

i.e.,

14

2

,

14

,

3 14

are the required direction cosines.

Ex.8

Find the scalar and vector components of the vector with initial point (2, 1) and terminal point (–5, 7).

Sol.

The vector joining the initial point (2, 1) and the terminal point (-5, 7) is = (- 5 - 2)ˆi + (7 - 1)ˆj = -7ˆi + 6ˆj The scalar components of the vectors are –7 and 6.

Ex.9

Find the unit vector in the direction of vector PQ , where P and Q are the points (1, 2, 3) and (4, 5, 6) respectively.

Sol.

The vector PQ = (4 - 1) ˆi + (5 - 2)ˆj + (6 - 3 )kˆ = 3ˆi + 3ˆj + 3kˆ

®

®

®

PQ = 3ˆi + 3ˆj + 3kˆ =

3 2 + 3 2 + 3 2 = 9 + 9 + 9 = 27 = 3 3

so, the unit vector in the direction of PQ =

1 3 3

(

)

. 3ˆi + 3ˆj + 3kˆ =

1 ˆ 1 ˆ 1 ˆ i+ j+ k 3 3 3

Ex.10 Find the direction cosines of the vector joining the points A(1, 2,–3) and B(–1,–2, 1), directed from A and B. Sol.

®

The vector AB = - 2ˆi - 4ˆj + 4kˆ The

direction

ratios

–2, –4 and 4 and, the direction cosines are

of

®

AB are,

therefore,

-1 -2 2 , , 3 3 3

Ex.11 Find the position vector of the mid-point of the vector joining the points P(2, 3, 4) and Q(4, 1, –2). Sol. The position vectors of the points P and Q are 2ˆi + 3ˆj + 4kˆ and 4ˆi + ˆj - 2kˆ The position vector of the mid-point of the vector PQ is

(

)

[

=

1 r r a+b 2

=

1é ˆ 6i + 4ˆj + 2kˆ ù = 3iˆ + 2ˆj + kˆ û 2ë

=

1 (2 + 4 )ˆi + (3 + 1)ˆj + (4 - 2)kˆ 2

]

r r b+a 1+ 1

Ex.12 Show that the points A, B and C with position vectors r r a = 3 iˆ - 4 ˆj - 4 kˆ , b = 2 iˆ - ˆj + kˆ

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VECTORS

7 B

r

and c = iˆ - 3 ˆj - 5 kˆ

A

C r b

r c

a

O

respectively form the vertices of a right-angled triangle. ®

Sol.

The vector AB = -ˆi + 3ˆj + 5 kˆ , ®

BC

®

= - ˆi - 2ˆj - 6kˆ ,

CA

= 2ˆi - ˆj + kˆ

®

Now, AB = AB = - ˆi + 3ˆj + 5kˆ = - 12 + 32 + 52 = 1 + 9 + 25 = 35 Þ AB 2 = 35 ®

BC = BC = - ˆi - 2ˆj - 6kˆ

(- 1)2 + (- 2)2 + (- 6)2

=

= 1 + 4 + 36 = 41 Þ BC 2 = 41

®

CA = CA = 2ˆi - ˆj + kˆ = 2 2 + (- 1)2 + 12 = 4 + 1 + 1 = 6

Þ

CA 2 = 6

Since BC2 = AB2 + CA 2; DABC is right- angled triangle, right-angled at A. Ex.13 Show that the vector iˆ + ˆj + kˆ is equally inclined to the axes OX, OY and OZ. Let a, b, g be the angles which the vector ˆi + ˆj + kˆ is inclined to the axes OX, OY and OZ respectively.. Then cos a = 1, cos b = 1 and cos g = 1 Þ a = b = g

Sol.

Ex.14 Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are ˆi + 2 ˆj - kˆ and - ˆi + ˆj + kˆ respectively, in the ratio 2 : 1. r

r

Sol.

Let the position vector of P be a = ˆi + 2ˆj - kˆ and of Q be b = -ˆi + ˆj + kˆ . (i) Let R divide the line joining the point P and Q in the ratio 2 :1 internally. Then, the position vector of R is r r 2b + a 2+1

=

(

1 r r 2b + a 3

()

r Q b

)

R r P (a)

[(

) (

)]

=

1 - 2ˆi + 2ˆj + 2kˆ + ˆi + 2ˆj - kˆ 3

=

1 (- 2 + 1)ˆi + (2 + 2) ˆj + (2 - 1)kˆ = 1 - ˆi + 4ˆj + kˆ 3 3

[

]

[

]

1 4 1 = - ˆi + ˆj + kˆ 3

3

(ii)

3

Let R divide the line joinin R

()

r Q b r P (a)

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VECTORS

8

the points P and Q in the ratio 2 :1 externally. Then the position r r r r 2b - a = 2b - a 2 -1 - 2ˆi + 2ˆj + 2kˆ - ˆi + 2ˆj - kˆ = - 3ˆi + 3kˆ

Vector of R is =

(

)(

)

Ex.15 Show that the points A(1, –2, –8), B(5, 0, –2) and C(11, 3, 7) are collinear and find the ratio in which B divides AC. Sol.

Here

®

AB = (5 - 1)ˆi + (0 + 2)ˆj + (- 2 + 8)kˆ = 4ˆi + 2ˆj + 6kˆ ®

and BC = (11 - 5 )ˆi + (3 - 0)ˆj + (7 + 2)kˆ = 6ˆi + 3ˆj + 9kˆ ®

®

4

2

6

The vectors AB and BC are collinear if = = 6 3 9

k A(1,-2,-8)

which is true. Hence, the given points are collinear. Let B divide AC in the ratio k : 1 Then the coordinates of B are :

1 B

C(11,3,7)

æ 11k + 1 3k - 2 7k - 8 ö , , ç ÷ è k +1 k +1 k +1 ø æ 11k + 1 3k - 2 7k - 8 ö , , ç ÷ è k +1 k +1 k +1 ø

Equating

which is true for k =

to (5, 0, 2), we have

11k + 1 k +1

=5;

3k - 2 k +1

= 0;

7k - 8 k +1

2 3

Hence, B divides AC internally in the ratio 2 : 3. Ex.16 Show that the following points are collinear: A( -2 ˆi + 3 ˆj + 5 kˆ ), B( ˆi + 2 ˆj + 3 kˆ ) and C ( 7 iˆ - kˆ ) Sol.

®

AB = (1 + 2)i + (2 - 3)ˆj + (3 - 5 )kˆ = 3ˆi - ˆj - 2kˆ

®

BC = (7 - 1)ˆi - 2ˆj + (- 1 - 3 )kˆ = 6ˆi - 2ˆj - 4kˆ

® ˆ ˆ ˆ AC = (7 + 2)ˆi - 3ˆj + (- 1 - 5 )kˆ = 9 i - 3 j - 6k

®

Further, AB = 9 + 1 + 4 = 14 ®

BC = 36 + 4 + 16 = 56 = 2 14 ®

AC = 81 + 9 + 36 = 126 = 3 14 ®

®

®

Therefore, | AC |=| AB | + | BC | Hence, the points A, B and C are collinear. r r r r r r r r r r Ex.17 If a, b, c are unit vectors such that ar + b + cr = 0 , find the value of a . b + b . c + c . a . r r r Sol. It is given that | a |=| b |=| c |= 1

r r r r Now, ar + b + cr = 0 gives a = -(b + c )

[ ( )] (

)

r r r r r r r rr (Q ar . ar = | ar |2 = 12 = 1) \ a . a = a. - b + c = - a.b + ac rr rr Þ a.b + a.c = -1 r r r r rr rr Þ a.b + c.a = -1 (Q a . c + c . a ) r r rr rr Similarly, b.cr + ar.b = -1 and c.a + b.c = -1 rr rr rr 3 rr rr rr a.b + b.c + c.a = Adding the three, we get 2 a.b + b.c + c.a = -3 Þ 2

(

)

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= –2

VECTORS 9 r r r r r r Ex.18 If a = 0 or b = 0 , then a . b = 0 . But the converse need not be true. Justify your answer with an example. r r r r r Sol. If a = 0 or b = 0, the ar.b = 0 r

r

But the converse is not necessarily true. For, take a = 2ˆi - ˆj + 3kˆ and b = 3ˆi + 3ˆj - kˆ r r Then ar.b = 0. because ar.b = 6 - 3 - 3 = 0 Yet. the vectors are non-zero vector. Ex.19 If the vertices A, B, C of a triangle ABC are (1, 2, 3), (–1, 0, 0), (0, 1, 2), respectively, then find Ð ABC. Sol.

®

Here,| BA |= (1 + 1)ˆi + (2 - 0 )ˆj + (3 - 0) kˆ

A(1, 2, 3)

= 2ˆi + 2ˆj + 3kˆ ®

Þ

| BA |= 22 + 22 + 32

4 + 4 + 9 = 17

B(–1, 0, 0)

C(0, 1, 2)

®

and | BC |= (0 + 1)ˆi + (1 - 0 )ˆj + (2 - 0 )kˆ = ˆi + ˆj + 2kˆ ® ®

®

Þ | BC |= 12 + 12 + 22 = 1 + 1 + 4 = 6 So,

( 2iˆ + 2jˆ + 3kˆ ).( ˆi + ˆj + 2kˆ ) =

®

17. 6 cos ÐABC 10

Þ 2 + 2 + 6 = 102 cos ÐABC Þ Þ

®

BA.BC =| BA | . | BC | .cos ÐABC

= cos ÐABC

102

æ 10 ö ÐABC = cos-1 ç ÷ è 102 ø

Ex.20 The scalar product of the vector ˆi + ˆj + kˆ with a unit vector along the sum of vectors 2 iˆ + 4 ˆj - 5 kˆ and liˆ + 2 ˆj + 3 kˆ is equal to one. Find the value of l . Sol.

r

r

ˆ bˆ = 2iˆ + 4ˆj - 5kˆ andc = l ˆi + 2jˆ + 3kˆ Let a = ˆi + ˆj + k,

Then b + c = (2ˆi + 4ˆj - 5kˆ ) + (l ˆi + 2ˆj + 3kˆ ) r

r

= ( 2 + l ) ˆi + (4 + 2)jˆ + ( -5 + 3)kˆ = (2 + l )ˆi + 6ˆj - 2kˆ

Þ Unit vector along r r r r b+c 1 b+c = r r = (2 + l )ˆi + 6ˆj - 2kˆ 2 |b+c| (2 + l ) + 36 + 4

[

2+l

=

(2 + l)

2

ˆi + + 40

6

(2 + l)

2

ˆj + 40

]

2kˆ

( 2 + l )2 + 40 r

Now, we are given that ar . unit vector along b + cr = 1 é

So, (ˆi + ˆj + kˆ ). êê Þ Þ

2+l

(2 + l )

2

ë

2+l

(2 + l )2 + 40

+ 40

+

2+l+6-2

(2 + l )2 + 40

Þ l+6=

ˆi +

6

(2 + l )2 + 40

6

(2 + l )

2

+ 40

-

j-

2

ù ú =1 (2 + l )2 + 40 úû 2kˆ

(2 + l )2 + 40

=1

=1

(2 + l )2 + 40

Þ (l + 6)2 = (2 + l )2 + 40 Þ l2 + 12l + 36 = l2 + 4l + 4 + 40 Þ 8l = 8 Þ l = 1

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VECTORS

10 Ex.21 Sol.

r r Let a = iˆ + 4 ˆj + 2 kˆ , b = 3 iˆ - 2 ˆj + 7 kˆ and r r c = 2 iˆ - ˆj + 4 kˆ . Find a vector d which is perpendicular to r Let d = a1ˆi + a2 ˆj + a3kˆ r r rr rr Since vector d is perpendicular to both ar and b , d.a = 0 and d.b = 0

r

r

r r

both a and b , and c . d = 15.

(a1ˆi + a2ˆj + a3kˆ )(. ˆi + 4ˆj + 2kˆ ) = 0

....(1)

a1 + 4a2 + 2a3 = 0

Þ

(a1ˆi + a2ˆj + a3kˆ ).(3iˆ - 2jˆ + 7kˆ ) = 0

3a1 - 2a2 + 4a3 = 0

.....(2)

r Also, crd = 0 Þ 2ˆi - ˆj + 4kˆ . a1ˆi + a2ˆj + a3kˆ = 15

(

)(

)

Þ 2a1 - a2 + 4a3 = 15 .....(3) Solving a1 , a2 and a3 from (1), (2) and (3),

we have a1 =

Ex.22 Sol. Ex.23 Sol.

160 3

5 3

, a2 = –

70 and a3 = -

3 1 Thus, the required vector is 160ˆi - 5ˆj - 70kˆ . 3 r r r r r r Show that ( a - b ) × ( a + b ) = 2 ( a ´ b ) r r r r r r r r r r r r r r r r r r r r Consider a - b ´ a + b = a - b ´ a + a - b ´ b = ar ´ ar - b ´ ar + ar ´ b - b ´ b = 0 + a ´ b + a ´ b - 0 = 2 a ´ b r Find l and m if ( 2 ˆi + 6 ˆj + 27 kˆ ) × ( ˆi + lˆj + mkˆ ) = 0 . r Given 2ˆi + 6ˆj + 27kˆ ´ i + lˆj + mkˆ = 0

(

)(

(

r d

(

) (

)

)(

)

(

)

( )

)

Þ 2ˆi ´ ˆi + 6ˆj ´ ˆi + 27kˆ ´ ˆi + 2l ˆi ´ j + 6lˆj ´ ˆj + 27lkˆ ´ ˆj r + 2mˆi ´ kˆ + 6mˆj ´ ˆj + 27mkˆ ´ kˆ = 0 r Þ 2 ´ 0 - 6i ´ ˆj + 27kˆ ´ ˆi + 2l ˆi ´ j + 6l ´ 0 - 27lˆj ´ kˆ - 2 μkˆ ´ ˆi + 6mˆj ´ kˆ + 27 ´ 0 = 0

Þ

-6kˆ + 27ˆj + 2lkˆ - 27l ˆi - 2mˆj + 6mˆi = 0

Þ

6m - 27l = 0; 27 - 2m = 0

Þ

(6m - 27l )i + (27 - 2m )j + (2l - 6 )kˆ = 0

and

2l - 6 = 0

Þ

l = 3, m =

27 2

27 Hence, l = 3 and m = . 2

r r

r

r

r

r

r

Ex.24 Given that a . b = 0 and a × b = 0 . What can you conclude about the vectors a and b ? r r Sol. Given ar . b = 0, we have | ar | | b | cos q = 0 r Either | ar | = 0, | b | = 0 or q = 90º r r r Given ar × b = 0 , we have | ar | | b | sin q = 0 r Either | ar | = 0, | b | = 0 or q = 0 r Hence, From (1) and (2) taken together, we have Either | ar | = 0 or | b | = 0 r

r

r

r

r

r

r

Ex.25 If either a = 0 or b = 0 , then a ´ b = 0 . Is the converse true ? Justify your answer with an example. Sol.

r

r

The converse not true. take any two non-zero collinear vectors, say, a = ˆi - ˆj + 2kˆ and b = 2ˆi - 2ˆj + 4kˆ Then,

r r r r r r r a ¹ 0 ; b ¹ 0 yet a ´ b = 0 ˆi ˆj r r a ´ b = 1 -1 2 -2

kˆ 2 4

r = ( -4 + 4 ) ˆi + ( 4 - 4 ) ˆj + ( -2 + 2 ) kˆ = 0

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VECTORS

11

UNSOLVED PROBLEMS EXERCISE – I r

r

r

r

Q.1

If a = 2ˆi + ˆj - kˆ, b = ˆi - ˆj + 2kˆ and c = ˆi - 2ˆj + kˆ such that ar + lb is perpendicular to cr , find the value of l.

Q.2

r r r Find the angles which vector ar makes with vectors b and cr , where a = 3ˆi + 2ˆj - kˆ, b = 2ˆi - ˆj - 3kˆ and r c = 2(ˆi - ˆj + kˆ ) . r

r

Q.3

Find the value of l for which the vectors a = lˆi + 2ˆj + 3kˆ and b = ˆi - 2ˆj - 3kˆ are (i) parallel, and (ii) perpendicular.

Q.4

Find the work done by the force F = 2ˆi + ˆj + 2kˆ in displacing an object from A(1, 2, –3) to B(3, 1, 2).

Q.5

Find the work done by the forces ˆi - ˆj + 2kˆ and 2ˆi + 3ˆj + kˆ in displacing an object from the origin to the point A(2, 1, 4).

Q.6

Find the projection of the vector ˆi + 2ˆj + 4kˆ on the vector 2ˆi + 4ˆj + kˆ .

Q.7

Find the component of the vector ˆi - ˆj + 2kˆ in the direction of 2ˆi + ˆj - kˆ .

Q.8

Show that the points whose position vectors are 4ˆi - 3ˆj + kˆ, 2ˆi - 4ˆj + 5kˆ and ˆi - ˆj from a right triangle.

Q.9

The adjacent sides of a triangle are 2ˆi - ˆj + 4kˆ and ˆi - 2ˆj - kˆ . Find the area of the triangle.

Q.10

Find the area of a parallelogram whose adjacent sides are ˆi + 2ˆj + 2kˆ and 2ˆi + ˆj - kˆ .

Q.11

If A(1, –1, 2), B(2, 3, 1) and C(3, 2, –1) are the vertices of DABC, find its area using the vector method.

Q.12

If ˆi + 2ˆj - kˆ and ˆi + ˆj - 2kˆ are the two diagonals of a parallelogram, find its area.

Q.13

If A(2, –1, 1), B(1, 2, 0), C(3, 2, 2) and D(4, –1, 3) are the vertices of a parallelogram ABCD, find its area.

r

r

Q.14 If ar , b and cr are the position vectors of the vertices of DABC, show that the area of DABC is

1 r |a 2

r r × b + b × cr + cr × ar | r

r

r

Q.15

Prove that ( ar – b ) × ( ar + b ) = 2 ar × b . Interpret the result geometrically..

Q.16

Prove that | ar × b | = a2 . b2 - (a .b)2 .

Q.17

r r r If | ar | = 4, | b | = 3 and | ar × b | = 8, find ar . b .

r

r

r

r r

Q.18 Find a unit vector perpendicular to both the vectors ˆi + 2ˆj - 3kˆ and 3ˆi - ˆj + 2kˆ . Q.19

Find a vector of magnitude 5 units in a direction perpendicular to both the vectors ˆi + ˆj - 2kˆ and 2ˆi - ˆj + kˆ .

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VECTORS

12 Q.20

r r r r Prove that ar × ( b + cr ) + b × ( cr + ar ) + cr × ( ar + b ) = 0 .

Q.21

For any vector ar , prove that | ar × ˆi |2 + | ar × ˆj |2 + | ar × kˆ |2 = 2| ar |2.

Q.22

Find a unit vector perpendicular to the plane of DABC where the position vectors of A, B and C are 2ˆi - ˆj + kˆ, ˆi + ˆj + 2kˆ

and 2ˆi + 3kˆ respectively..

Q.23

Find the angles of DABC when the vertices are A(1, 2, –1), B(2, –1, 1) and C(1, 1, –2).

Q.24

r r r r For any vector rr , prove that r = ( r . ˆi )ˆi + ( r . ˆj )ˆj + ( r .kˆ )kˆ.

Q.25

If aˆ and bˆ are unit vectors inclined at an angle q then prove that sin = | aˆ – bˆ |. 2 2

q

1

r Q.26 If ar , b and cr are three mutually perpendicular vectors of equal magnitude, prove that the vector ar r

r

+ b + cr is equally inclined with the vectors ar , b and cr . r r r r r r r If ar = 3ˆi - ˆj and b = 2ˆi + ˆj - 3kˆ , express b in the form b = b1 + b2 , where b1 is parallel to ar and b2 is perpendicular to ar .

Q.27

r

r

r

r

r

r

r

r

Q.28

For any two vectors a and b , prove that | a + b |2 + | a – b |2 = 2(| a |2 + | b |2).

Q.29

r r r If ar + b + cr = 0 , | ar | = 3, | b | = 7 and | cr | = 5, find the angle between ar and cr .

Q.30

r r r r r If ar and b are two vectors such that | ar + b | = | ar | then prove that 2a + b is perpendicular to b .

ANSWER KEY EXERCISE – 1 (UNSOLVED PROBLEMS) 2. 60º, 90º 6. 11.

3. (i) –1 (ii) 13

2 21 3 107 2

18. ±

7.

-1

9.

6

12. 11 sq units

sq units

( ˆi - 11ˆj - 7kˆ )

19. ±

171 æ

4. 13 units

ö

5. 20 units 3 14 2

sq units

13. 6 2 sq units

5( ˆi + 5ˆj + 3kˆ ) 35

22. ± æ

10. 5 2 sq units 17. ± 4 5

3ˆi + 2ˆj - kˆ 14 ö

1 1 3 1 1 3 –1 æ 13 ö –1 –1 23. ÐA = cos çç 2 7 ÷÷ , ÐB = cos ç 14 ÷ , ÐC = cos çç 2 7 ÷÷ 27. b1 = ˆi - ˆj ; b2 = ˆi + ˆj - 3kˆ è ø 2 2 2 2 è ø è ø 29. 60º r

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r

VECTORS

13

BOARD PROBLES EXERCISE – II Q.1

r

r

Find l if a = 4ˆi - ˆj + kˆ and b = l ˆi - 2ˆj + 2kˆ are perpendicular to each other..

[C.B.S.E. 2000]

r r a = 3ˆi + ˆj - 2kˆ and b = 2ˆi + 3ˆj - kˆ .

Q.2

Find a unit vector perpendicular to both

Q.3

r Define a × b and prove that a ´ b a ´ b = (a . b) tan q , where q is the angle between r r r r

r

[C.B.S.E. 2000]

r r

r

r

the vectors a and b . Q.4

If three vectors

[C.B.S.E. 2000]

r r r r r r r r r r r r r a , b and c are such that a + b + c = 0 , prove that a ´ b = b ´ c = c ´ a

[C.B.S.E. 2001] r

r b = i - 2k

r r 2b ´ a

Q.5

If a = 4ˆi + 3ˆj + kˆ and

Q.6

Find a vector whose magnitude is 3 units and which is perpendicular to the following r

r

find

.

[C.B.S.E. 2001]

r

r

two vectors a and b : a = 3ˆi + ˆj - 4kˆ and b = 6ˆi + 5ˆj - 2kˆ

[C.B.S.E. 2001]

®

®

Q.7

In D OAB, OA = 3ˆi + 2ˆj - k and OB = ˆi + 3ˆj + k . Find the area of the triangle.

[C.B.S.E. 2002]

Q.8

r r r Prove that (ar ´ b )2 =| ar |2 .| b |2 -(ar . b)2 .

[C.B.S.E. 2002]

Q.9

For any two vectors a and b , show that (1 + | a |2 ) r

r

r

r ) (1 + | b | ) = (1 - a.b r 2

r 2

r r r r + | a + b + a ´ b |2

(

)

[C.B.S.E. 2002] Q.10

r r If a , b and cr are position vectors of points A, B and c, then prove that r r r r r r a ´ b + b ´ c + c ´ a is a vector perpendicular to the plane of D ABC.

(

Q.11 Q.12

)

[C.B.S.E. 2002]

Find the value of l so that the two vectors 2ˆi + 3ˆj - kˆ and. 4ˆi + 6ˆj + lkˆ are (i) Parallel (ii) Perpendicular to each other [C.B.S.E. 2003] Find the unit vector perpendicular to the plane ABC where the position vectors of A, B and C are 2ˆi - ˆj + kˆ, ˆi + ˆj + 2k and 2ˆi + 3kˆ respectively.. r

r

r

r

[C.B.S.E. 2004] r

r

Q.13

If a = 5ˆi - ˆj - 3kˆ and b = ˆi + 3ˆj - 5kˆ , then show that vectors a + b and a - b are orthogonal. [C.B.S.E. 2004]

Q.14

Show that the points whose position vectors are a = 4ˆi - 3ˆj + kˆ. bˆ = 2ˆi - 4ˆj + 5kˆ

r

r

and c = ˆi - ˆj form a right angled triangle. Q.15 Q.16 Q.17

(

Q.18

[C.B.S.E. 2005]

r r r r r Let a = ˆi - ˆj, b = 3 j - kˆ . Find a vector d which is perpendicular to both a and b r r and c . d =1. [C.B.S.E. 2005] r Express the vector a = 5ˆi - 2ˆj + 5kˆ as sum of two vectors such that one is parallel r r [C.B.S.E. 2005] to the vector b = 3 ˆi + kˆ and the other is perpendicular to b . r r r If a,b and c are mutually perpendicular vectors of equal magnitude, show that they are r r r equally inclined to the vector a + b + c . [C.B.S.E. 2006] r r r r r r If a = ˆi + 2ˆj - 3kˆ and b = 3ˆi - ˆj + 2kˆ , show that a + b and a - b are perpendicular to

)

each other. Q.19

[C.B.S.E. 2006]

Find the angle between the vectors

r r a+b

r r and a - b

r where a = 2ˆi - ˆj + 3kˆ

and

r b = 3ˆi - ˆj - 2kˆ

[C.B.S.E. 2006] Q.20

If

r a = ˆi + ˆj + kˆ

and

r b = ˆj - kˆ

find a vector

r r r r c such that a ´ c = b

and

r r a.c =3

www.thinkiit.in

.

[C.B.S.E. 2007]

VECTORS

14 Q.21

Find the projection of

r r r r r b + c on a where a = 2ˆi - 2ˆj + kˆ ,b = ˆi + 2ˆj - 2kˆ

and

r c = 2ˆi - ˆj + 4kˆ

[C.B.S.E. 2007]

r r r r r r r Three vectors a , b and c satisfy the condition a + b + c = 0 . Find the value of r r r rr rr rr a.b + b.c + c.a if a = 1, b = 4 and c = 2 .

Q.22

Q.23

Find a vector of magnitude 5 units. perpendicular to each of the vectors

(

r r a+b

[C.B.S.E. 2008]

) and (ar - br )

r

r

where a = ˆi + ˆj + kˆ and b = ˆi + 2ˆj + 3kˆ

[C.B.S.E. 2008]

If ˆi + ˆj + kˆ , 2ˆi + 5ˆj . 3ˆi + 2ˆj - 3kˆ and ˆi - 6ˆj - kˆ are the position vectors of the points A, B, C

Q.24

® ® ® ® . Deduce that AB are collinear.. and D, find the angle between AB and CD and CD [C.B.S.E. 2008]

r r r r r r r r If a + b + c = 0 and a = 3, b = 5 , and c = 7, show that angle between ar and b is 60º.

Q.25

[C.B.S.E. 2008] Q.26

The scalar product of the vector ˆi + ˆj + kˆ with a unit vector along the sum of vectors l ˆi + 2ˆj + 3kˆ 2ˆi + 4ˆj + 5kˆ

and is equal to one. Find the value of l . [C.B.S.E. 2009] r r r r r r If P is a unit vector and (x - p ) (x + p ) = 80 , then find x . [C.B.S.E. 2009] r r r r r Find the projection of a on b if a . b = 8 and b = 2ˆi + 6ˆj + 3kˆ [C.B.S.E. 2009] r r r If a = ˆi + ˆj + kˆ , b = 4ˆi - 2ˆj + 3kˆ and c = ˆi - 2ˆj + kˆ , find a vector of magnitude 6 units r r r which is parallel to the vector 2a - b + 3c . [C.B.S.E. 2010] r r r ˆ r Let a = i + 4ˆj + 2kˆ, b = 3ˆi - 2ˆj + 7kˆ and c = 2ˆi - ˆj + 4kˆ . Find a vector d which is r r r r perpendicular to both a and b and c . d = 18. [C.B.S.E. 2010]

Q.27 Q.28 Q.29 Q.30 Q.31

Using vectors, find the area of the triangle with vertices A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5). [C.B.S.E. 2011]

Q.32

Let a = i + 4 j+ 2 k , b = 3 i - 2 j+ 7 k and c = 2 i - j + 4 k . Find a vector p which is

®

^

^

^

®

^

^

^

^

^

®

^

® ®

®

®

®

perpendicular to both a and b and p . c = 18. ®

^

^

^

®

^

^

[C.B.S.E. 2012]

^

®

®

®

®

If a = i - j+ 7 k and b = 5 i - j + l k , then find the value of l, so that a + b and a - b are

Q.32

perpendicular vectors

[C.B.S.E. 2013]

ANSWER KEY EXERCISE – 2 (BOARD PROBLEMS) 1. –1

1

2.

5 3

(5 ˆi - ˆj + 7kˆ )

6. 2ˆi - 2ˆj + kˆ

7.

3 2

10 sq units

10. 60 27. 6 14

11. (i) –2 (ii)

26 1

12.

14

22. –

21 2

( 3 ˆi + 2ˆj - kˆ 23. –

31. 1 61 2

5

ˆ 6 i

15. +

10 6

32.

1 ( ˆi + ˆj + 3kˆ 4

ˆj

–

5

ˆ 6 k

®

16. 6ˆi + 2kˆ ; - ˆi - 2ˆj + 3kˆ

)

24. 180º 26. 1 27. 9 28. ^

^

8 7

19.

p 2

20.

5ˆ 2ˆ 2 ˆ i+ j+ k 3 3 3

21. 2

29. 2ˆi - 4ˆj + 4kˆ 30. 64ˆi - 2ˆj - 28kˆ

^

p = 64 i - 2 j - 28 k

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