If a → , b → , c → are mutually perpendicular, show that a → · b → × c → 2 = a 2 b 2 c 2 .

Given that a → , b → , c → are mutually perpendicular. a → · b → × c → 2 = a → · b → c → sin 90 ° n ^ 2 ∵ b → ⊥ c → n ^ is perpendicular to the plane containing both b → , c → = b c a → · n ^ 2 = b c a → n ^ cos 0 2 ∵ n ^ ∥ a → = a b c 2 = a 2 b 2 c 2 Hence, proved.

Show that a → + b → b → + c → c → + a → = 2 a → b → c → .

LHS = a → + b → b → + c → c → + a → = a → + b → · b → + c → × c → + a → = a → + b → · b → × c → + b → × a → + c → × c → + c → × a → = a → + b → · b → × c → + b → × a → + c → × a → = a → · b → × c → + b → · b → × c → + a → · b → × a → + b → · b → × a → + a → · c → × a → + b → · c → × a → = a → · b → × c → + b → · c → × a → = a → b → c → + b → c → a → = a → b → c → + a → b → c → = 2 a → b → c → = RHS Hence, proved.

Show that a → × b → b → × c → c → × a → = a → b → c → 2 .

LHS = a → × b → b → × c → c → × a → = a → × b → × b → × c → · c → × a → We know that a → × b → × c → = a → · c → b → - b → · c → a → Now, a → × b → × b → × c → = a → · b → × c → b → - b → · b → × c → a → = b → a → b → c → - a → b → b → c → = b → a → b → c → - 0 = b → a → b → c → Now, we have a → × b → × b → × c → · c → × a → = b → a → b → c → · c → × a → = a → b → c → b → · c → × a → = a → b → c → b → c → a → = a → b → c → a → b → c → = a → b → c → 2 = RHS Hence, proved.

For a → = i ^ + j ^ , b → = - i ^ + 2 k ^ , c → = j ^ + k ^ , obtain a → × b → × c → and also verify the formula a → × b → × c → = a → · c → b → - a → · b → c →

Given a → = i ^ + j ^ , b → = - i ^ + 2 k ^ , c → = j ^ + k ^ Now, b → × c → = i ^ j ^ k ^ - 1 0 2 0 1 1 = 0 - 2 i ^ - - 1 - 0 j ^ + - 1 - 0 k ^ = - 2 i ^ + j ^ - k ^ And a → × b → × c → = i ^ j ^ k ^ 1 1 0 - 2 1 - 1 = - 1 - 0 i ^ - - 1 - 0 j ^ + 1 + 2 k ^ = - i ^ + j ^ + 3 k ^ Now, a → · c → = i ^ + j ^ · j ^ + k ^ = 1 a → · b → = i ^ + j ^ · - i ^ + 2 k ^ = - 1 And a → · c → b → - a → · b → c → = 1 - i ^ + 2 k ^ + 1 j ^ + k ^ = - i ^ + j ^ + 3 k ^ Hence, proved.

Prove that a → × b → × c → + b → × c → × a → + c → × a → × b → = 0 → and hence prove that a → × b → × c → , b → × c → × a → , c → × a → × b → are coplanar.

LHS = a → × b → × c → + b → × c → × a → + c → × a → × b → = a → · c → b → - a → · b → c → + b → · a → c → - b → · c → a → + c → · b → a → - c → · a → b → = a → · c → b → - a → · b → c → + a → · b → c → - b → · c → a → + b → · c → a → - a → · c → b → ∵ c → · d → = d → · c → = 0 → = RHS Hence, proved. We know that three vectors are coplanar if p → q → r → = 0 Let p → = a → × b → × c → , q → = b → × c → × a → , r → = c → × a → × b → We have a → × b → × c → + b → × c → × a → + c → × a → × b → = 0 → ⇒ p → + q → + r → = 0 → ⇒ p → = - q → + r → ⇒ p → × q → = - q → + r → × q → ⇒ p → × q → = - q → × q → - r → × q → ⇒ p → × q → = 0 - r → × q → ⇒ p → × q → · r → = - r → × q → · r → ⇒ p → q → r → = 0 . Therefore, a → × b → × c → , b → × c → × a → , c → × a → × b → are coplanar.

If a ^ , b ^ , c ^ are unit vectors and a ^ × b ^ × c ^ = 1 2 b ^ , find angles that a ^ makes with b ^ and c ^ , where b ^ , c ^ are not parallel.

Given that a ^ , b ^ , c ^ are unit vectors ⇒ a ^ = b ^ = c ^ = 1 We have a ^ × b ^ × c ^ = 1 2 b ^ ⇒ a ^ · c ^ b ^ - a ^ · b ^ c ^ = 1 2 b ^ ⇒ a ^ · c ^ b ^ - a ^ · b ^ c ^ × b ^ = 1 2 b ^ × b ^ ⇒ a ^ · c ^ b ^ × b ^ - a ^ · b ^ c ^ × b ^ = 1 2 b ^ × b ^ ⇒ 0 - a ^ · b ^ c ^ × b ^ = 0 ⇒ a ^ · b ^ c ^ × b ^ = 0 ⇒ a ^ · b ^ = 0 as b ^ , c ^ are not parallel c ^ × b ^ ≠ 0 ⇒ a ^ b ^ cos θ = 0 ⇒ cos θ = 0 ⇒ θ = π 2 And a ^ × b ^ × c ^ = 1 2 b ^ ⇒ a ^ · c ^ b ^ - a ^ · b ^ c ^ = 1 2 b ^ ⇒ a ^ · c ^ b ^ - a ^ · b ^ c ^ × c ^ = 1 2 b ^ × c ^ ⇒ a ^ · c ^ b ^ × c ^ - a ^ · b ^ c ^ × c ^ = 1 2 b ^ × c ^ ⇒ a ^ · c ^ b ^ × c ^ - 0 = 1 2 b ^ × c ^ ⇒ a ^ · c ^ = 1 2 ⇒ a ^ b ^ cos θ = 1 2 ⇒ cos θ = 1 2 ⇒ θ = π 3 .