Find the unit vector in the direction of the vector a → = j ^ + i ^ + 2 k ^ .

The unit vector a ^ in the direction of vector a → = i ^ + j ^ + 2 k ^ is given by a ^ = a → | a | | a → | = 1 2 + 1 2 + 2 2 = 1 + 1 + 4 = 6 ∴ a ^ = a → | a → | = i ^ + j ^ + 2 k ^ 6 = 1 6 i ^ + 1 6 j ^ + 2 6 k ^

Find the unit vector in the direction of the vector a → = j ^ + i ^ + 2 k ^ .

The unit vector a ^ in the direction of vector a → = i ^ + j ^ + 2 k ^ is given by a ^ = a → | a | | a → | = 1 2 + 1 2 + 2 2 = 1 + 1 + 4 = 6 ∴ a ^ = a → | a → | = i ^ + j ^ + 2 k ^ 6 = 1 6 i ^ + 1 6 j ^ + 2 6 k ^

Let a → = i ^ + 4 j ^ + 2 k ^ b → = 3 i ^ - 2 j ^ + 7 k ^ c → = 2 i ^ - j ^ + 4 k ^ Find a vector d → which is perpendicular to both a → and b → , c → · d → = 15 .

If two vectors a → and b → are given in component form as a 1 i ^ + a 2 j ^ + a 3 k ^ and b 1 i ^ + b 2 j ^ + b 3 k ^ , then their scalar product is given as a → · b → = a 1 b 1 + a 2 b 2 + a 3 b 3 . Given, a → = i ^ + 4 j ^ + 2 k ^ , b → = 3 i ^ - 2 j ^ + 7 k ^ and c → = 2 i ^ - j ^ + 4 k ^ Let d → = d 1 i ^ + d 2 j ^ + d 3 k ^ Two vectors are perpendicular implies that dot product of the vectors is zero. d → · a → = 0 ⇒ d 1 + 4 d 2 + 2 d 3 = 0 .............(i) d → · b → = 0 ⇒ 3 d 1 - 2 d 2 + 7 d 3 = 0 ...........(ii) c → · d → = 15 ⇒ 2 d 1 - d 2 + 4 d 3 = 15 ........(iii) From equation (i) and (ii) we have d 1 28 + 4 = d 2 6 - 7 = d 3 - 2 - 12 = k ⇒ d 1 32 = d 2 - 1 = d 3 - 14 = k ⇒ d 1 = 32 k , d 2 = - k , and d 3 = - 14 k Substituting the values of d 1 , d 2 , and d 3 in equation (iii) 2 × 32 k - - k + 4 - 14 k = 15 ⇒ 64 k + k - 56 k = 15 ⇒ 9 k = 15 ⇒ k = 5 3 ∴ d 1 = 32 × 5 3 = 160 3 , d 2 = - 5 3 a n d d 3 = - 14 × 5 3 = - 70 3 Hence, d → = 160 3 i ^ - 5 3 j ^ - 70 3 k ^ = 1 3 ( 160 i ^ - 5 j ^ - 70 k ^ )

If a → , b → , c → are unit vectors such that a → + b → + c → = 0 , find the value of a → . b → + b → . c → + c → . a → .

Given, a → , b → and c → are unit vectors. So, a → = b → = c → = 1 Also given that a → + b → + c → = 0 → So, a → + b → + c → = 0 → = 0 Now, a → + b → + c → 2 = a → + b → + c → ⋅ a → + b → + c → = a → · a → + a → · b → + a → · c → + b → · a → + b → · b → + b → · c → + c → · a → + c → · b → + c → · c → [Using distributivity of scalar product] Now, using the commutative property of dot product i.e. a → · b → = b → · a → ⇒ a → + b → + c → 2 = a → · a → + a → · b → + c → · a → + a → · b → + b → · b → + b → · c → + c → · a → + b → · c → + c → · c → ⇒ 0 = a → · a → + b → · b → + c → · c → + 2 a → · b → + b → · c → + c → · a → ⇒ 0 = a → 2 + b → 2 + c → 2 + 2 a → · b → + b → · c → + c → · a → [ Using a → · a → = a → 2 ] ⇒ 0 = 1 + 1 + 1 + 2 a → · b → + b → · c → + c → · a → ⇒ 0 = 3 + 2 a → · b → + b → · c → + c → · a → ⇒ a → ⋅ b → + b → ⋅ c → + c → ⋅ a → = - 3 2

Find the unit vector perpendicular to both a → and b → where a → = 3 i ^ + j ^ - 2 k ^ and b → = 2 i ^ + 3 j ^ - k ^ .

Given a → = 3 i ^ + j ^ - 2 k ^ and b → = 2 i ^ + 3 j ^ - k ^ The unit vector perpendicular to both the vectors a → and b → is ± a → × b → a → × b → . Cross product of vectors a → and b → is a → × b → = i ^ j ^ k ^ 3 1 - 2 2 3 - 1 ⇒ a → × b → = i ^ - 1 + 6 - j ^ - 3 + 4 + k ^ 9 - 2 ⇒ a → × b → = 5 i ^ - j ^ + 7 k ^ The unit vector perpendicular to both the vectors a → and b → is ± a → × b → a → × b → = ± 5 i ^ - j ^ + 7 k ^ 5 2 + - 1 2 + 7 2 = ± 5 i ^ - j ^ + 7 k ^ 25 + 1 + 49 = ± 5 i ^ - j ^ + 7 k ^ 75 = ± 5 i ^ - j ^ + 7 k ^ 5 3 Hence, unit vector perpendicular to a → = 3 i ^ + j ^ - 2 k ^ and b → = 2 i ^ + 3 j ^ - k ^ is ± 5 i ^ - j ^ + 7 k ^ 5 3 .

If a → = 5 i ^ - j ^ + 7 k ^ and b → = i ^ - j ^ - λ k ^ , find the value of λ for which a → + b → and a → - b → are perpendicular to each other.

Given : a → = 5 i ^ - j ^ + 7 k ^ and b → = i ^ - j ^ - λ k ^ ⇒ a → + b → = 5 i ^ - j ^ + 7 k ^ + i ^ - j ^ - λ k ^ ⇒ a → + b → = 6 ı ^ - 2 ȷ ^ + ( 7 - λ ) k ^ Similarly a → - b → = 5 i ^ - j ^ + 7 k ^ - i ^ - j ^ - λ k ^ ⇒ a → - b → = 4 i ^ + 0 j ^ + ( 7 + λ ) k ^ We know that if two vectors are perpendicular then their dot product is zero. Given that a → + b → and a → - b → are perpendicular to each other. ( a → + b → ) . ( a → - b → ) = 0 ⇔ [ 6 ı ^ - 2 ȷ ^ + ( 7 - λ ) k ^ ] . [ 4 i ^ + 0 j ^ + ( 7 + λ ) k ^ ] = 0 ⇒ 6 × ( 4 ) + ( - 2 ) × 0 + ( 7 - λ ) ( 7 + λ ) = 0 ⇒ 24 + 7 2 - λ 2 = 0 ⇒ 73 - λ 2 = 0 ⇒ λ 2 = 73 ⇒ λ = ± 73 Hence required value of λ is ± 73 .

Let a → = i ^ + 4 j ^ + 2 k ^ b → = 3 i ^ - 2 j ^ + 7 k ^ c → = 2 i ^ - j ^ + 4 k ^ Find a vector d → which is perpendicular to both a → and b → , c → · d → = 15 .

If two vectors a → and b → are given in component form as a 1 i ^ + a 2 j ^ + a 3 k ^ and b 1 i ^ + b 2 j ^ + b 3 k ^ , then their scalar product is given as a → · b → = a 1 b 1 + a 2 b 2 + a 3 b 3 . Given, a → = i ^ + 4 j ^ + 2 k ^ , b → = 3 i ^ - 2 j ^ + 7 k ^ and c → = 2 i ^ - j ^ + 4 k ^ Let d → = d 1 i ^ + d 2 j ^ + d 3 k ^ Two vectors are perpendicular implies that dot product of the vectors is zero. d → · a → = 0 ⇒ d 1 + 4 d 2 + 2 d 3 = 0 .............(i) d → · b → = 0 ⇒ 3 d 1 - 2 d 2 + 7 d 3 = 0 ...........(ii) c → · d → = 15 ⇒ 2 d 1 - d 2 + 4 d 3 = 15 ........(iii) From equation (i) and (ii) we have d 1 28 + 4 = d 2 6 - 7 = d 3 - 2 - 12 = k ⇒ d 1 32 = d 2 - 1 = d 3 - 14 = k ⇒ d 1 = 32 k , d 2 = - k , and d 3 = - 14 k Substituting the values of d 1 , d 2 , and d 3 in equation (iii) 2 × 32 k - - k + 4 - 14 k = 15 ⇒ 64 k + k - 56 k = 15 ⇒ 9 k = 15 ⇒ k = 5 3 ∴ d 1 = 32 × 5 3 = 160 3 , d 2 = - 5 3 a n d d 3 = - 14 × 5 3 = - 70 3 Hence, d → = 160 3 i ^ - 5 3 j ^ - 70 3 k ^ = 1 3 ( 160 i ^ - 5 j ^ - 70 k ^ )

Prove that the points A ( 2 , 0 , - 3 ) , B ( 3 , 2 , - 1 ) and C ( 1 , - 2 , - 5 ) are collinear.

Points A , B and C are collinear if they lie on a same line. If A , B , C are collinear then A B → = k B C → , where k is a real number. A B → = O B → - O A → A B → = 3 i ^ + 2 j ^ - k ^ - ( 2 i ^ + 0 j ^ - 3 k ^ ) A B → = j ^ + 2 j ^ + 2 k ^ Now, B C → = O C → - O B → B C → = i ^ - 2 j ^ - 5 k ^ - ( 3 i ^ + 2 j ^ - k ^ ) B C → = - 2 i ^ - 4 j ^ - 4 k ^ B C → = - 2 i ^ + 2 j ˙ + 2 k B C → = - 2 A B → Therefore, the given points are collinear

Using vectors, find the area of ∆ A B C whose vertices are A 1 , 2 , 3 , B 2 , 5 , - 1 and C - 1 , 1 , 2

Given vertices of ∆ A B C are A 1 , 2 , 3 , B 2 , 5 , - 1 and C - 1 , 1 , 2 . Now, through the given vertices, we can find the vectors representing the adjacent sides, A B , and A C of ∆ A B C . A B → = O B → - O A → = 2 i ^ + 5 j ^ - k ^ - i ^ + 2 j ^ + 3 k ^ = 2 - 1 i ^ + 5 - 2 j ^ + - 1 - 3 k ^ = i ^ + 3 j ^ - 4 k ^ And, A C → = O C → - O A → = - i ^ + j ^ + 2 k ^ - i ^ + 2 j ^ + 3 k ^ = - 2 i ^ - j ^ - k ^ Now, we know that the area of the triangle = a → × b → 2 , where a → , and b → are the vectors representing the adjacent sides of the triangle. A B → × A C → = i ^ j ^ k ^ 1 3 - 4 - 2 - 1 - 1 = i ^ - 3 - 4 - j ^ - 1 - 8 + k ^ - 1 + 6 = - 7 i ^ + 9 j ^ + 5 k ^ A B → × A C → = - 7 2 + 9 2 + 5 2 = 49 + 81 + 25 = 155 Hence, area of the triangle= 155 2 = 12 . 449899598 2 = 6 . 225 square units.