Let a⃗=î+α ĵ+β k̂, α, β ∈ ℝ. Let a vector b⃗ be such that the angle between a⃗ and b⃗ is π/4 and |b⃗|²=6. If a⃗ · b⃗=3 √2, then the value of (α²+β²)|a⃗ × b⃗|² is equal to

Solution

Maths sir · 3 steps

Dot product and angle give you |a⃗|, and then α²+β² = |a⃗|² - 1.

1. Length of a. Here |b⃗| = √6 and cos π/4 = 1/√2. So a⃗·b⃗ = |a⃗|√6·1/√2 = √3 |a⃗|. Setting this equal to 3√2 gives |a⃗| = √6, so |a⃗|² = 6.
2. Get alpha and beta. From a⃗ = î + αĵ + βk̂, |a⃗|² = 1 + α² + β² = 6. So α² + β² = 5.

   |a⃗×b⃗|² = |a⃗|²|b⃗|² - (a⃗·b⃗)² = 6 × 6 - (3√2)² = 36 - 18 = 18

3. So. (α²+β²) |a⃗×b⃗|² = 5 × 18 = 90, which is option (b).

The idea to keep. Remember |a⃗×b⃗|² + (a⃗·b⃗)² = |a⃗|²|b⃗|². Once you know the dot product and both lengths, the cross product needs no components at all.

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