Chapter 10 – Vector Algebra

Find the unit vector in the direction of the sum of the vectors a = 2 iˆ − ˆj + 2 kˆ and b = – iˆ + ˆj + 3 kˆ . Solution Let c denote the sum of a and b . We have c = (2 iˆ − ˆj + 2 kˆ) + (−iˆ + ˆj + 3 kˆ) = iˆ + 5 kˆ Now | c | = 12 + 52 = 26 . VECTOR ALGEBRA 207 Thus, the required unit vector is c 26 ( 26 ) c = c = 1 i + 5k = 1 i + 5 k .

Find a vector of magnitude 11 in the direction opposite to that of PQ , where P and Q are the points (1, 3, 2) and (–1, 0, 8), respetively. Solution The vector with initial point P (1, 3, 2) and terminal point Q (–1, 0, 8) is given by PQ = (– 1 – 1) iˆ + (0 – 3) ĵ + (8 – 2) k̂ = – 2 iˆ – 3 ĵ + 6 k̂ Thus Q P = – P Q = 2 iˆ + 3 ˆj − 6 kˆ ⇒ | QP | = 22 + 32 + (–6) 2 = 4 + 9 + 36 = 49 = 7 Therefore, unit vector in the direction of Q P is given by Q P = 2 iˆ + 3 ˆj − 6 kˆ Q P = |Q P| 7 Hence, the required vector of magnitude 11 in direction of Q P is  2 iˆ + 3 ˆj − 6 kˆ  22 ˆ 33 ˆ 66 ˆ = 11 

Find the position vector of a point R which divides the line joining the two points P and Q with position vectors O P = 2 a + b and O Q = a – 2 b , respectively,, in the ratio 1:2,

• (i)internally and
• (ii)externally. Solution (i) The position vector of the point R dividing the join of P and Q internally in the ratio 1:2 is given by 2( 2 a + b ) + 1( a – 2 b ) 5 a OR = = . 1+ 2 3 208 MATHEMATICS (ii) The position vector of the point R′ dividing the join of P and Q in the ratio 1 : 2 externally is given by 2(2a + b ) − 1( a − 2 b) O R′ = = 3a + 4 b . 2 −1

If the points (–1, –1, 2), (2, m, 5) and (3,11, 6) are collinear, find the value of m. Solution Let the given points be A (–1, –1, 2), B (2, m, 5) and C (3, 11, 6). Then A B = (2 + 1) iˆ + (m + 1) ˆj + (5 – 2) kˆ = 3 iˆ + (m + 1) ˆj + 3kˆ and A C = (3 + 1) iˆ + (11 + 1) ˆj + (6 − 2) kˆ = 4 iˆ + 12 ˆj + 4kˆ . Since A, B, C, are collinear, we have A B = λ A C , i.e., ( 3 iˆ + (m + 1) ˆj + 3kˆ) = λ (4iˆ +12 ˆj + 4kˆ) ⇒ 3 = 4 λ and m + 1 = 12 λ Therefore m = 8. π

Find a vector r of magnitude 3 2 units which makes an angle of and π with y and z - axes, respectively.. π 1 π Solution Here m = cos = and n = cos = 0. 4 2 2 Therefore, l2 + m2 + n2 = 1 gives l2 + +0 =1 ⇒ l = ± VECTOR ALGEBRA 209 Hence, the required vector r = 3 2 (l iˆ + m ˆj + n kˆ) is given by 1 ˆ 1 ˆ r = 3 2 (± i+ j + 0 kˆ) = r = ± 3 iˆ + 3 ˆj . 2 2

If a = 2 iˆ − ˆj + kˆ, b = iˆ + ĵ – 2kˆ and c = iˆ + 3jˆ – k̂ , find λ such that a is perpendicular to b + c . Solution We have λ b + c = λ ( iˆ + ĵ – 2 k̂ ) + ( iˆ + 3 ĵ – k̂ ) = ( λ + 1) iˆ + (λ + 3 ) ĵ – (2λ + 1) k̂ Since a ⊥ (λ b + c ), a .(λ b + c ) = 0 ⇒ (2 iˆ – ĵ + k̂ ) . [( λ + 1) iˆ + (λ + 3) ĵ – (2λ + 1) k̂ ] = 0 ⇒ 2 (λ + 1) – (λ + 3) – (2λ + 1) = 0 ⇒ λ = – 2.

Find all vectors of magnitude 10 3 that are perpendicular to the plane of iˆ + 2 ˆj + kˆ and −iˆ + 3 ˆj + 4kˆ . Solution Let a = iˆ + 2 ˆj + kˆ and b = −iˆ + 3 ˆj + 4kˆ . Then iˆ ˆj kˆ a × b = 1 2 1 = iˆ(8 − 3) − ˆj (4 + 1) + kˆ (3 + 2) = 5 iˆ – 5 ĵ + 5 k̂ –1 3 4 ⇒ a × b = (5)2 + ( −5)2 + (5)2 = 3(5)2 = 5 3 . 210 MATHEMATICS Therefore, unit vector perpendicular to the plane of a and b is given by a × b 5iˆ − 5 ˆj + 5kˆ = a ×b 5 3 Hence, vectors of magnitude of 10 3 that are perpendicular to plane of a and b  5iˆ − 5 ˆj + 5kˆ  are ± 10 3   , i.e., ± 10(iˆ − ˆj + kˆ) .  5 3 

Using vectors, prove that cos (A – B) = cosA cosB + sinA sinB. and OQ Solution Let OP be unit vectors making angles A and B, respectively, with positive direction of x-axis. Then ∠QOP = A – B [Fig. 10.1] = OM We know OP = ON + NQ = iˆ cos B + ˆj sin B. + MP = iˆ cos A + ˆj sin A and OQ OQ By definition OP. = cos ( A-B) OP OQ = cos (A – B) ... (1) ( OP = 1= OQ ) In terms of components, we have OQ OP. = (iˆ cos A + ˆj sin A).(iˆ cos B + ˆj sin B) = cosA cosB + sinA sinB ... (2) From (1) and (2), we get cos (A – B) = cosA cosB + sinA sinB. VECTOR ALGEBRA 211 sin A sin B sin C

Prove that in a ∆ ABC, = = , where a, b, c represent the a b c magnitudes of the sides opposite to vertices A, B, C, respectively. Solution Let the three sides of the triangle BC, CA and AB be represented by a , b and c , respectively [Fig. 10.2]. We have a + b + c = 0 . i.e., a + b = − c which pre cross multiplying by a , and post cross multiplying by b , gives a ×b = c × a and a ×b =b ×c respectively. Therefore, a ×b =b ×c =c ×a ⇒ a ×b = b ×c = c ×a ⇒ a b sin ( π – C) = b c sin ( π – A) = c a sin ( π – B) ⇒ ab sin C = bc sinA = ca sinB Dividing by abc, we get sin C sin A sin B sin A sin B sin C = = i.e. = = c a b a b c

The magnitude of the vector 6iˆ + 2 ˆj + 3kˆ is 212 MATHEMATICS

• (A)5
• (B)7
• (C)12
• (D)1 Solution (B) is the correct answer.

The position vector of the point which divides the join of points with position vectors a + b and 2a − b in the ratio 1 : 2 is 3a + 2b 5a − b 4a + b

• (A)
• (B)a
• (C)
• (D)3 3 3 Solution (D) is the correct answer. Applying section formula the position vector of the required point is 2( a + b ) + 1(2a − b ) 4a + b = 2 +1 3

The vector with initial point P (2, –3, 5) and terminal point Q(3, –4, 7) is

• (A)iˆ − ˆj + 2kˆ
• (B)5iˆ − 7 ˆj + 12kˆ
• (C)−iˆ + ˆj − 2kˆ
• (D)None of these Solution (A) is the correct answer.

The angle between the vectors iˆ − ˆj and ĵ − kˆ is π 2π −π 5π

• (A)
• (B)
• (C)
• (D)3 3 3 6 a .b Solution (B) is the correct answer. Apply the formula cosθ = a . b .

The value of λ for which the two vectors 2iˆ − ˆj + 2kˆ and 3iˆ + λˆj + kˆ are perpendicular is

• (A)2
• (B)4
• (C)6
• (D)8 Solution (D) is the correct answer. VECTOR ALGEBRA 213

The area of the parallelogram whose adjacent sides are iˆ + kˆ and 2iˆ + ˆj + kˆ is

• (A)2
• (B)3
• (C)3
• (D)4 Solution (B) is the correct answer. Area of the parallelogram whose adjacent sides are a and b is a ×bˆ .

If a = 8, b = 3 and a × b = 12 , then value of a . b is

• (A)6 3
• (B)8 3
• (C)12 3
• (D)None of these Solution (C) is the correct answer. Using the formula a × b = a . b |sinθ|, we get π θ= ± . 3 Therefore, a .b = a . b cos θ = 8 × 3 × = 12 3 .

The 2 vectors ˆj + kˆ and 3iˆ − ˆj + 4kˆ represents the two sides AB and AC, respectively of a ∆ABC. The length of the median through A is

The projection of vector a = 2iˆ − ˆj + kˆ along b = iˆ + 2 ˆj + 2kˆ is 214 MATHEMATICS 2 1

• (A)
• (B)
• (C)2
• (D)6 3 3 Solution (A) is the correct answer. Projection of a vector a on b is a .b (2iˆ − ˆj + kˆ).(iˆ + 2 ˆj + 2kˆ) 2 = = . b 1+ 4 + 4 3

If a and b are unit vectors, then what is the angle between a and b for 3a − b to be a unit vector?

• (A)30°
• (B)45°
• (C)60°
• (D)90° Solution (A) is the correct answer. We have ( 3a − b )2 = 3a 2 + b 2 − 2 3 a.b 3 3 ⇒ a .b = ⇒ cosθ = ⇒ θ = 30°. 2 2

The unit vector perpendicular to the vectors iˆ − ˆj and iˆ + ˆj forming a right handed system is iˆ − ˆj iˆ + ˆj

• (A)k̂
• (B)– k̂
• (C)
• (D)2 2 (iˆ − ˆj ) × (iˆ + ˆj ) 2kˆ ˆ ( ) ( ) 2 =k . Solution (A) is the correct answer. Required unit vector is iˆ − ˆj × iˆ + ˆj =

If a = 3 and – 1≤ k ≤ 2 , then ka lies in the interval

• (A)[0, 6]
• (B)[– 3, 6]
• (C)[ 3, 6]
• (D)[1, 2] VECTOR ALGEBRA 215 Solution (A) is the correct answer. The smallest value of ka will exist at numerically smallest value of k, i.e., at k = 0, which gives ka = k a = 0 × 3 = 0 The numerically greatest value of k is 2 at which ka = 6 .