Find the angle between two vectors a → and b → , if a → × b → = a → · b → .

Let θ be the angle between a → and b → . Given that, a → × b → = a → . b → ⇒ a → b → sin θ = a → b → cos θ ⇒ sin θ = cos θ ⇒ tan θ = 1 ⇒ θ = π 4 Hence, the required angle between them is π 4 .

If a → × b → = b → × c → ≠ 0 , then show that a → + c → = m b → , where m is any scalar.

Given that: a → × b → = b → × c → We have to prove that: a → + c → = m b → ⇒ a → × b → = - c → × b → ⇒ a → × b → + c → × b → = 0 ⇒ ( a → + c → ) × b → = 0 (Using right distributive property) Thus, a → + c → is parallel to b → . Hence, a → + c → = m b → , for some scalar m .

What inference can you draw if a → × b → = 0 → and a → · b → = 0 .

Given that, a → × b → = 0 → ⇒ a → b → sin θ = 0 ⇒ a → = 0 o r b → = 0 ∴ a → ‖ b → Also, a → . b → = 0 ⇒ a → b → cos θ = 0 ⇒ a → = 0 → o r b → = 0 → o r , a → ⊥ b → But a → cannot be both perpendicular as well as parallel to b → . Thus, a → = 0 , b → = 0

If a → , b → , c → are three unit vectors such that a → × b → = c → , b → × c → = a → , c → × a → = b → . Show that a → , b → , c → form an orthonormal right handed triad of unit vectors.

Given that: a → × b → = c → , b → × c → = a → and c → × a → = b → … . . ( 1 ) Now | a → × b → | = | c → | = 1 ( ∵ c → is a unit vector ) | b → × c → | = | a → | = 1 ( ∵ a → is a unit vector ) | c → × a → | = | b → | = 1 ( ∵ b → is a unit vector ) ∴ | a → × b → | = | b → × c → | = | c → × a → | = 1 … . . ( 2 ) From ( 1 ) and ( 2 ) , we conclude that Thus, a → , b → and c → form an orthonormal right-handed triad of unit vectors Hence, Proved.

If a , b , c are the lengths of sides, B C , C A and A B of a triangle A B C , prove that B C → + C A → + A B → = 0 → and deduce that a sin A = b sin B = c sin C

Let: B C → = a → , C A → = b → , A B → = c → Then, | a → | = a | b → | = b ( ∵ Length is always positive) c → = c Given that: B C → + C A → + A B → = 0 → ⇒ a → + b → + c → = 0 → a → × ( a → + b → + c → ) = a → × 0 → a → × a → + a → × b → + a → × c → = 0 → 0 → + a → × b → - c → × a → = 0 → a → × b → = c → × a → | a → | | b → | sin C = | c → | | a → | sin B a · b sin C = c · a sin B Dividing both sides by a b c , we get sin C c = sin B b . . . . . 1 Again, B C → + C A → + A B → = 0 → a → + b → + c → = 0 → b → × ( a → + b → + c → ) = b → × 0 → b → × a → + b → × b → + b → × c → = 0 → - a → × b → + 0 → + b → × c → = 0 → a → × b → = b → × c → | a → | | b → | sin C = | b → | | c → | sin A a b sin C = b c sin A Dividing both sides by a b c , we get ⇒ sin C c = sin A a . . . . . 2 From ( 1 ) and ( 2 ) , we get sin A a = sin B b = sin C c ⇒ a sin A = b sin B = c sin C Hence, proved.

If a → = i ^ - 2 j ^ + 3 k ^ , and b → = 2 i ^ + 3 j ^ - 5 k ^ , then find a → × b → . Verify that a → and a → × b → are perpendicular to each other.

Given that: a → = i ^ - 2 j ^ + 3 k ^ and b → = 2 i ^ + 3 j ^ - 5 k ^ So, a → × b → = i ^ j ^ k ^ 1 - 2 3 2 3 - 5 = i ^ + 11 j ^ + 7 k ^ Now, a → · ( a → × b → ) = 1 - 22 + 21 = 0 Thus, a → is perpendicular to a → × b → Hence, verified.