Let b⃗ = î + ĵ + λ k̂, λ ∈ R. If a⃗ is a vector such that a⃗ × b⃗ = 13î - ĵ - 4k̂ and a⃗ . b⃗ + 21 = 0, then ( b⃗ - a⃗ ). ( k̂ - ĵ ) + ( b⃗ + a⃗ ). ( î - k̂ ) is equal to _____________.

Solution

Maths sir · 4 steps

Use a⃗ × b⃗ ⊥ b⃗ to fix λ, then rebuild a⃗ with a triple product.

1. Find lambda. a⃗ × b⃗ is perpendicular to b⃗, so (13î - ĵ - 4k̂) · (î + ĵ + λ k̂) = 0. That gives 13 - 1 - 4λ = 0, so λ = 3 and b⃗ = î + ĵ + 3k̂, with |b⃗|² = 11.

   b⃗ × (a⃗ × b⃗) = |b⃗|² a⃗ - (a⃗ · b⃗) b⃗ ⇒ a⃗ = (b⃗ × (a⃗ × b⃗) + (a⃗ · b⃗) b⃗)/|b⃗|²

2. Rebuild a. b⃗ × (13î - ĵ - 4k̂) = -î + 43ĵ - 14k̂ and (a⃗ · b⃗)b⃗ = -21î - 21ĵ - 63k̂. Adding and dividing by 11 gives a⃗ = -2î + 2ĵ - 7k̂. Check: a⃗ · b⃗ = -2 + 2 - 21 = -21.
3. Evaluate both parts. b⃗ - a⃗ = 3î - ĵ + 10k̂, so its dot product with k̂ - ĵ is 1 + 10 = 11. b⃗ + a⃗ = -î + 3ĵ - 4k̂, so its dot product with î - k̂ is -1 + 4 = 3.
4. So. The expression equals 11 + 3 = 14.

The idea to keep. Know a⃗ · b⃗ and a⃗ × b⃗? Cross b⃗ with a⃗ × b⃗ to recover a⃗. And use (a⃗ × b⃗) ⊥ b⃗ first to find any unknown.

Answer from the NTA final answer key. Solution by PrepWiser.