Two vectors A⃗ and B⃗ have equal magnitudes. If magnitude of A⃗ + B⃗ is equal to two times the magnitude of A⃗ - B⃗, then the angle between A⃗ and B⃗ will be :

Solution

Physics sir · 3 steps

Square both magnitudes. With equal sizes, the ratio leaves you with a simple equation in cos θ.

1. Write both magnitudes. Let |A⃗| = |B⃗| = a and let θ be the angle between them. Then |A⃗+B⃗|² = 2a²(1+cos θ) and |A⃗-B⃗|² = 2a²(1-cos θ).
2. Square the condition. |A⃗+B⃗| = 2|A⃗-B⃗| becomes |A⃗+B⃗|² = 4|A⃗-B⃗|². The 2a² cancels on both sides.

   (1+cos θ)/(1-cos θ) = 4 ⇒ 1+cos θ = 4-4 cos θ ⇒ cos θ = 3/5

3. So. θ = cos⁻¹(3/5), which is option (c). In sine form this angle is sin⁻¹(4/5), not sin⁻¹(3/5).

The idea to keep. For equal vectors, |A⃗±B⃗|² = 2a²(1±cos θ). Always square the given ratio before you compare, so a factor of 2 becomes 4.

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