Let a⃗=a₁ î+a₂ ĵ+a₃ k̂ and b⃗=b₁ î+b₂ ĵ+b₃ k̂ be two vectors such that |a⃗|=1, a⃗ · b⃗=2 and |b⃗|=4. If c⃗=2(a⃗ × b⃗)-3 b⃗, then the angle between b⃗ and c⃗ is equal to:

Solution

Maths sir · 4 steps

Dot b⃗ into c⃗: the cross-product part is perpendicular to b⃗ and vanishes.

1. Find b dot c. The vector a⃗ × b⃗ is perpendicular to b⃗, so (a⃗ × b⃗) · b⃗ = 0. That leaves b⃗ · c⃗ = 2(0) - 3|b⃗|² = -3 × 16 = -48.
2. Length of cross product. Use |a⃗ × b⃗|² = |a⃗|²|b⃗|² - (a⃗ · b⃗)² = 1 × 16 - 4 = 12.
3. Size of c. The two parts of c⃗ are perpendicular, so the cross term is zero. Then |c⃗|² = 4 × 12 + 9 × 16 = 48 + 144 = 192, so |c⃗| = 8√3.

   cos θ = (b⃗ · c⃗)/(|b⃗| |c⃗|) = -48/(4 × 8√3) = -3/2√3 = -√3/2

4. So. The angle between b⃗ and c⃗ is cos⁻¹(-√3/2), which is option (d).

The idea to keep. Whenever a vector has an a⃗ × b⃗ part, dot it with a⃗ or b⃗ and that part vanishes. Use |a⃗ × b⃗|² = a²b² - (a⃗ · b⃗)² for its length.

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