If a → = 2 i ^ + 5 j ^ - 7 k ^ , b → = - 3 i ^ + 4 j ^ + k ^ and c → = i ^ - 2 j ^ - 3 k ^ , compute ( a → × b → ) × c → and a → × ( b → × c → ) and verify that these are not equal.

Given that: a → = 2 i ^ + 5 j ^ - 7 k ^ , b → = - 3 i ^ + 4 j ^ + k ^ and c → = i ^ - 2 j ^ - 3 k ^ ∴ a → × b → = i ^ j ^ k ^ 2 5 - 7 - 3 4 1 = ( 5 + 28 ) i ^ - ( 2 - 21 ) j ^ + ( 8 + 15 ) k ^ = 33 i ^ + 19 j ^ + 23 k ^ ⇒ ( a → × b → ) × c → = i ^ j ^ k ^ 33 19 23 1 - 2 - 3 = ( - 57 + 46 ) i ^ - ( - 99 - 23 ) j ^ + ( - 66 - 19 ) k ^ ⇒ ( a → × b → ) × c → = - 11 i ^ + 122 j ^ - 85 k ^ . . . . . ( i ) ∴ b → × c → = i ^ j ^ k ^ - 3 4 1 1 - 2 - 3 = ( - 12 + 2 ) i ^ - ( 9 - 1 ) j ^ + ( 6 - 4 ) k ^ = - 10 i ^ - 8 j ^ + 2 k ^ ⇒ a → × ( b → × c → ) = i ^ j ^ k ^ 2 5 - 7 - 10 - 8 2 = ( 10 - 56 ) i ^ - ( 4 - 70 ) j ^ + ( - 16 + 50 ) k ^ ⇒ a → × ( b → × c → ) = - 46 i ^ + 66 j ^ + 34 k ^ . . . . . ( i i ) From ( 1 ) and ( 2 ) , we get ( a → × b → ) × c → ≠ a → × ( b → × c → ) Hence, verified.

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 .