Let a⃗=î+ĵ+k̂, b⃗=2 î+2 ĵ+k̂ and d⃗=a⃗ × b⃗. If c⃗ is a vector such that a⃗ · c⃗=|c⃗|, |c⃗-2 a⃗|²=8 and the angle between d⃗ and c⃗ is π/4, then |10-3 b⃗ · c⃗|+|d⃗ × c⃗|² is equal to _________.

Solution

Maths sir · 5 steps

Expand the square first: it pins down |c⃗|, and then every term falls into place.

1. Find d. d⃗ = a⃗ × b⃗ = (1-2)î - (1-2)ĵ + (2-2)k̂ = -î + ĵ, so |d⃗| = √2.
2. Fix the length. Expand |c⃗-2a⃗|² = |c⃗|² - 4 a⃗·c⃗ + 4|a⃗|². Use a⃗·c⃗ = |c⃗| and |a⃗|² = 3 to get |c⃗|² - 4|c⃗| + 12 = 8. That is (|c⃗|-2)² = 0, so |c⃗| = 2 and a⃗·c⃗ = 2.
3. Cross product term. The angle between d⃗ and c⃗ is π/4, so use |d⃗×c⃗| = |d⃗||c⃗|sin π/4 and square it.

   |d⃗×c⃗|² = |d⃗|²|c⃗|²sin² π/4 = 2 × 4 × 1/2 = 4

4. Find b dot c. Let c⃗ = xî + yĵ + zk̂. From d⃗·c⃗ = √2 × 2 × 1/√2 = 2 we get y = x + 2, and x + y + z = 2 gives z = -2x. Then |c⃗|² = 4 gives 6x² + 4x = 0, so x = 0 or x = -2/3, and b⃗·c⃗ = 2x + 2y + z = 2x + 4 is 4 or 8/3. In both cases |10 - 3 b⃗·c⃗| = 2.
5. So. The value is 2 + 4 = 6.

The common slip: 4. Forgetting to square |d⃗×c⃗| = 2 gives 2 + 2 = 4, not 6.

The idea to keep. When a dot product equals a length, expand the given |u⃗-v⃗|² and look for a perfect square. It usually fixes the magnitude in one line.

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