If a → and b → are perpendicular vectors show that a → + b → 2 = a → - b → 2 , a → + b → 2 = a → + b → · a → + b → so does a → - b → 2

Given that a → and b → are perpendicular vectors. ⇒ a → · b → = b → · a → = 0 Now, a → + b → 2 = a → + b → · a → + b → = a → · a → + a → · b → + b → · a → + b → · b → = a → 2 + b → 2 . . . . 1 And a → - b → 2 = a → - b → · a → - b → = a → · a → - a → · b → - b → · a → + b → · b → = a → 2 + b → 2 . . . . 2 ∴ a → + b → 2 = a → - b → 2 Hence, proved.

If a → and b → are perpendicular vectors show that a → + b → 2 = a → - b → 2 , a → + b → 2 = a → + b → · a → + b → so does a → - b → 2

Given that a → and b → are perpendicular vectors. ⇒ a → · b → = b → · a → = 0 Now, a → + b → 2 = a → + b → · a → + b → = a → · a → + a → · b → + b → · a → + b → · b → = a → 2 + b → 2 . . . . 1 And a → - b → 2 = a → - b → · a → - b → = a → · a → - a → · b → - b → · a → + b → · b → = a → 2 + b → 2 . . . . 2 ∴ a → + b → 2 = a → - b → 2 Hence, proved.

Prove that two non-zero vectors are perpendicular if a → + b → 2 = a → 2 + b → 2 .

Given that a → + b → 2 = a → 2 + b → 2 We know that a → + b → 2 = a → + b → · a → + b → ⇒ a → · a → + a → · b → + b → · a → + b → · b → = a → 2 + b → 2 ⇒ a → 2 + b → 2 + 2 a → b → cos θ = a → 2 + b → 2 , θ is angle between a → , b → ⇒ 2 a → b → cos θ = 0 ⇒ 2 a → · b → = 0 ⇒ a → · b → = 0 a → , b → are two non-zero vectors , θ = π 2 We know that two vectors are perpendicular if dot product between them is zero. Therefore, a → and b → are perpendicular vectors.

If a → , b → , c → are mutually perpendicular vectors of equal magnitude, show that a → + b → + c → is equally inclined with a → , b → , c → .

Given that a → , b → , c → have equal magnitude. ⇒ a → = b → = c → And a → , b → , c → are mutually perpendicular vectors ⇒ a → · b → = b → · c → = c → · a → = 0 Now, a → + b → + c → · a → = a → + b → + c → a → cos α , where α is the angle between a → + b → + c → and a → ⇒ a → · a → + b → · a → + c → · a → = a → + b → + c → a → cos α ⇒ a → 2 = a → + b → + c → a → cos α ⇒ cos α = a → a → + b → + c → Now, a → + b → + c → · b → = a → + b → + c → b → cos β , where β is the angle between a → + b → + c → and b → ⇒ a → · b → + b → · b → + c → · b → = a → + b → + c → b → cos β ⇒ b → 2 = a → + b → + c → b → cos β ⇒ cos β = b → a → + b → + c → Now, a → + b → + c → · c → = a → + b → + c → c → cos γ , where γ is the angle between a → + b → + c → and c → ⇒ a → · c → + b → · c → + c → · c → = a → + b → + c → c → cos γ ⇒ c → 2 = a → + b → + c → c → cos γ ⇒ cos γ = c → a → + b → + c → ∴ cos α = cos β = cos γ ∵ a → = b → = c → ⇒ α = β = γ Hence, proved.