Let a⃗=2 î+ĵ-k̂, b⃗=((a⃗ ×(î+ĵ)) × î) × î. Then the square of the projection of a⃗ on b⃗ is:

Solution

Maths sir · 4 steps

Do not expand all three cross products. The triple product rule gives b⃗ in one line.

1. First cross product. Let v⃗ = a⃗ × (î+ĵ). Expanding the determinant of (2, 1, -1) and (1, 1, 0) gives v⃗ = î - ĵ + k̂.
2. Triple product rule. Use (p⃗ × q⃗) × r⃗ = (p⃗·r⃗)q⃗ - (q⃗·r⃗)p⃗ with q⃗ = r⃗ = î. Since î·î = 1, the two outer crosses collapse.

   b⃗ = (v⃗×î)×î = (v⃗·î) î - v⃗ = î - (î - ĵ + k̂) = ĵ - k̂

3. Find the projection. a⃗·b⃗ = 0 + 1 + 1 = 2 and |b⃗| = √2. So the projection of a⃗ on b⃗ is (a⃗·b⃗)/|b⃗| = 2/√2 = √2.
4. So. The square of the projection is (√2)² = 2, which is option (c).

The idea to keep. When a cross product is crossed again, use (p⃗×q⃗)×r⃗ = (p⃗·r⃗)q⃗ - (q⃗·r⃗)p⃗. Projection on b⃗ divides by |b⃗|.

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