Question

If are mutually perpendicular vectors of equal magnitudes, show that the vector is equally inclined to and .

Answer 1

It is given that a → , b → , c → are mutually perpendicular vectors of equal magnitudes. So,

a → ⋅ b → = b → ⋅ c → = c → ⋅ a → =0 (i)

Apply dot product on ( a → + b → + c → )with a → .

( a → + b → + c → )⋅ a → =| ( a → + b → + c → ) || a → |cosα

Where αis the angle between ( a → + b → + c → )and a → .

a → ⋅ a → + b → ⋅ a → + c → ⋅ a → =| a → + b → + c → || a → |cosα | a → | 2 +0+0=| a → + b → + c → || a → |cosα cosα= | a → | | a → + b → + c → |

Apply dot product on ( a → + b → + c → )with b →

( a → + b → + c → )⋅ b → =| ( a → + b → + c → ) || b → |cosβ

Where βis the angle between ( a → + b → + c → )and b → .

a → ⋅ b → + b → ⋅ b → + c → ⋅ b → =| a → + b → + c → || b → |cosβ 0+ | b → | 2 +0=| a → + b → + c → || b → |cosβ cosβ= | b → | | a → + b → + c → |

Apply dot product on ( a → + b → + c → )with c → .

( a → + b → + c → )⋅ c → =| ( a → + b → + c → ) || c → |cosγ

Where γis the angle between ( a → + b → + c → )and c → .

a → ⋅ c → + b → ⋅ c → + c → ⋅ c → =| a → + b → + c → || c → |cosγ 0+0+ | c → | 2 =| a → + b → + c → || c → |cosγ cosγ= | c → | | a → + b → + c → |

So,

cosα= | a → | | a → + b → + c → | , cosβ= | b → | | a → + b → + c → | , cosγ= | c → | | a → + b → + c → |

As | a → |=| b → |=| c → |, so,

cosα=cosβ=cosγ α=β=γ

Thus, ( a → + b → + c → )is equally inclined to a → , b → , c → .