Let a⃗ and b⃗ be two vectors such that |a⃗+b⃗|²=|a⃗|²+2|b⃗|², a⃗ · b⃗=3 and |a⃗ × b⃗|²=75. Then |a⃗|² is equal to __________.

Solution

Maths sir · 4 steps

The first condition hands you |b⃗|²; then Lagrange's identity gives |a⃗|² in one line.

1. Expand the square. |a⃗+b⃗|² = |a⃗|² + 2 a⃗·b⃗ + |b⃗|². Set this equal to |a⃗|² + 2|b⃗|² and cancel |a⃗|² to get 2 a⃗·b⃗ = |b⃗|².
2. Find b squared. With a⃗·b⃗ = 3, this gives |b⃗|² = 2 × 3 = 6.

   |a⃗×b⃗|² = |a⃗|²|b⃗|² - (a⃗·b⃗)²

3. Put in numbers. 75 = 6|a⃗|² - 3² = 6|a⃗|² - 9, so 6|a⃗|² = 84.
4. So. |a⃗|² = 84/6 = 14.

The common slip: 28. Expanding |a⃗+b⃗|² with only one a⃗·b⃗ term gives |b⃗|² = 3, and then |a⃗|² = 28 instead of 14.

The idea to keep. Whenever both a⃗·b⃗ and |a⃗×b⃗| appear, reach for |a⃗×b⃗|² + (a⃗·b⃗)² = |a⃗|²|b⃗|². It links them without any angle.

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