If a 1 → , a 2 → , a 3 → and b 1 → , b 2 → , b 3 → be two sets of non-coplanar vectors, such that a p → ⋅ b q → = 0 i f p ≠ q 4 i f p = q for p = 1 , 2 , 3 and q = 1 , 2 , 3 , then the value of a 1 → 2 a 2 → 3 a 3 → b 1 → + b 2 → b 2 → + b 3 → b 3 → + b 1 → is equal to

Step-by-step solution

a 1 → 2 a 2 → 3 a 3 → b 1 → + b 2 → b 2 → + b 3 → b 3 → + b 1 → = 6 a 1 → a 2 → a 3 → 2 b 1 → b 2 → b 3 → = 12 a 1 → ⋅ b 1 → a 1 → ⋅ b 2 → a 1 → ⋅ b 3 → a 2 → ⋅ b 1 → a 2 → ⋅ b 2 → a 2 → ⋅ b 3 → a 3 → ⋅ b 1 → a 3 → ⋅ b 2 → a 3 → ⋅ b 3 → = 12 4 0 0 0 4 0 0 0 4 = 12 × 64 = 768