Let a⃗ and b⃗ be two vectors
such that | 2a⃗ + 3b⃗ | = | 3a⃗ + b⃗ | and the angle between a⃗ and b⃗ is 60°. If 1/8a⃗ is a unit vector, then | b⃗ | is equal to :

Solution

Maths sir · 4 steps

Square both sides so the moduli turn into dot products, then use |a⃗| = 8 and cos 60°.

1. Read off a. 1/8a⃗ is a unit vector, so |a⃗| = 8. With the angle 60°, a⃗ · b⃗ = 8|b⃗| × 1/2 = 4|b⃗|.
2. Square both sides. Use |u⃗|² = u⃗ · u⃗. The left side becomes 4|a⃗|² + 12 a⃗ · b⃗ + 9|b⃗|² and the right side becomes 9|a⃗|² + 6 a⃗ · b⃗ + |b⃗|². Moving everything to one side gives the equation below.

   8|b⃗|² + 6 a⃗ · b⃗ - 5|a⃗|² = 0

3. Put in numbers. Write x = |b⃗|: 8x² + 24x - 320 = 0, so x² + 3x - 40 = 0, which factorises as (x + 8)(x - 5) = 0.
4. So. A magnitude cannot be negative, so x = -8 is rejected and |b⃗| = 5, option (c).

The idea to keep. When a question gives equal moduli of vector sums, square them: |u⃗|² = u⃗ · u⃗ turns everything into |a⃗|², |b⃗|² and a⃗ · b⃗.

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