Prove that a → a 2 - b → b 2 2 = a → - b → a b 2 .

We have to prove that a → a 2 - b → b 2 2 = a → - b → a b 2 . Consider LHS and proceed to check if it is the same as RHS. a → a 2 - b → b 2 2 = a → a 2 2 + b → b 2 2 - 2 a → a 2 · b → b 2 = a → 2 a 4 + b → 2 b 4 - 2 a → · b → a 2 b 2 ∵ a → 2 = a → 2 = a 2 , b → 2 = b → 2 = b 2 = a 2 a 4 + b 2 b 4 - 2 a → · b → a 2 b 2 = 1 a 2 + 1 b 2 - 2 a → · b → a 2 b 2 Taking LCM, we have = b 2 a 2 b 2 + a 2 b 2 a 2 - 2 a → · b → a 2 b 2 = a 2 + b 2 - 2 a → · b → a 2 b 2 = a → - b → 2 a 2 b 2 = a → - b → a b 2 = RHS Hence, proved.

Prove that a → a 2 - b → b 2 2 = a → - b → a b 2 .

We have to prove that a → a 2 - b → b 2 2 = a → - b → a b 2 . Consider LHS and proceed to check if it is the same as RHS. a → a 2 - b → b 2 2 = a → a 2 2 + b → b 2 2 - 2 a → a 2 · b → b 2 = a → 2 a 4 + b → 2 b 4 - 2 a → · b → a 2 b 2 ∵ a → 2 = a → 2 = a 2 , b → 2 = b → 2 = b 2 = a 2 a 4 + b 2 b 4 - 2 a → · b → a 2 b 2 = 1 a 2 + 1 b 2 - 2 a → · b → a 2 b 2 Taking LCM, we have = b 2 a 2 b 2 + a 2 b 2 a 2 - 2 a → · b → a 2 b 2 = a 2 + b 2 - 2 a → · b → a 2 b 2 = a → - b → 2 a 2 b 2 = a → - b → a b 2 = RHS Hence, proved.

Given that p → = a → + b → , q → = a → - b → and a → = b → , show that p → · q → = 0 .

We have to prove that p → · q → = 0 . So, substituting the data given in the question, p → · q → = ( a → + b → ) · ( a → - b → ) = a → · ( a → - b → ) + b → · ( a → - b → ) = a → · a → - a → · b → + b → · a → - b → · b → = a → 2 - b → 2 = a → 2 - b → 2 ∵ a → = b → ⇒ p → · q → = 0 Hence proved.

Find a → - b → if two vectors are such that a → = 2 , b → = 3 and a → · b → = 4 .

The details given in the question are a → = 2 , b → = 3 and a → · b → = 4 . Now, to find the value of a → - b → consider, a → - b → = a → - b → 2 = a → 2 + b → 2 - 2 a → · b → Substituting the values, a → - b → = 2 2 + 3 2 - 2 × 4 = 5 Therefore, a → - b → = 5 .

If a → is a unit vector and x → - a ^ · x → + a ^ = 12 , then find x → .

The condition given in the question is x → - a ^ · x → + a ^ = 12 and a → is a unit vector. Starting with the first condition, x → - a ^ · x → + a ^ = 12 ⇒ x → · x → + a ^ - a ^ · x → + a ^ = 12 ⇒ x → · x → + x → · a → - a ^ · x → - a → · a ^ = 12 ⇒ x → 2 - a ^ 2 = 12 ∵ x → 2 = x → 2 , a → 2 = a → 2 ⇒ x → 2 - a ^ 2 = 12 ⇒ x → 2 - 1 2 = 12 ⇒ x → 2 = 12 + 1 ⇒ x → 2 = 13 ⇒ x → = 13

Find a → and b → , if a → + b → · a → - b → = 8 and a → = 8 b → .

There are two conditions a → + b → · a → - b → = 8 and a → = 8 b → given in the question. Starting with the first one, a → + b → · a → - b → = 8 ⇒ a → · a → - b → + b → · a → - b → = 8 ⇒ a → · a → - a → · b → + b → · a → - b → · b → = 8 ⇒ a → 2 - b → 2 = 8 ∵ a → 2 = a → 2 , b → 2 = b → 2 ∴ a → 2 - b → 2 = 8 Now apply the second condition, a → = 8 b → in the above result, 8 b → 2 - b → 2 = 8 ⇒ 64 b → 2 - b → 2 = 8 ⇒ 63 b → 2 = 8 ⇒ b → 2 = 8 63 ⇒ b → = 8 63 ⇒ b → = 2 2 3 7 Now substituting the value of b → in the second condition, a → = 8 b → ⇒ a → = 8 × 2 2 3 7 ⇒ a → = 16 2 3 7

Show that a → b → + b → a → is perpendicular to a → b → - b → a → for any two non-zero vectors a → and b → .

Two vectors are perpendicular if their scalar product is zero. So, checking the condition, a → b → + b → a → · ( a → b → - b → a → ) = a → b → · ( a → b → - b → a → ) + b → a → · ( a → b → - b → a → ) = a → b → · a → b → - a → b → · b → a → + b → a → · a → b → - b → a → · b → a → = ( a → b → ) 2 - ( b → a → ) 2 = a → 2 b → 2 - b → 2 a → 2 ∵ a → 2 = a → 2 , b → 2 = b → 2 = a → 2 b → 2 - b → 2 a → 2 = 0 Hence, given vectors are perpendicular to each other.

If a → = 1 , b → = 2 , c → = 3 and a → + b → + c → = 0 → , then show that a → · b → + b → · c → + c → · a → = - 7 .

In the question, it is given that a → + b → + c → = 0 → ⇒ a → + b → + c → = 0 . Taking its square, a → + b → + c → 2 = 0 Using the formula ( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 ( a b + b c + c a ) and applying it, a → + b → + c → 2 = 0 ⇒ a → 2 + b → 2 + c → 2 + 2 ( a → · b → + b → · c → + c → · a → ) = 0 ∵ a → 2 = a → 2 , b → 2 = b → 2 , c → 2 = c → 2 ⇒ a → 2 + b → 2 + c → 2 + 2 ( a → · b → + b → · c → + c → · a → ) = 0 Substituting a → = 1 , b → = 2 and c → = 3 in the above equation, ⇒ 1 2 + 2 2 + 3 2 + 2 ( a → · b → + b → · c → + c → · a → ) = 0 ⇒ 1 + 4 + 9 + 2 ( a → · b → + b → · c → + c → · a → ) = 0 ⇒ 2 ( a → · b → + b → · c → + c → · a → ) = - 14 ⇒ a → · b → + b → · c → + c → · a → = - 7 Hence proved.