Two vectors P⃗ and Q⃗ have equal magnitudes. If the magnitude of P⃗+Q⃗ is n times the magnitude of P⃗-Q⃗, then angle between P⃗ and Q⃗ is :

Solution

Physics sir · 4 steps

Square both magnitudes first; with equal vectors the ratio collapses into one equation for cos θ.

1. Write both magnitudes. Let |P⃗| = |Q⃗| = P and the angle between them be θ. Then |P⃗+Q⃗|² = 2P²(1+cos θ) and |P⃗-Q⃗|² = 2P²(1-cos θ).
2. Square the condition. The condition |P⃗+Q⃗| = n |P⃗-Q⃗| becomes |P⃗+Q⃗|² = n² |P⃗-Q⃗|² once you square it. The factor 2P² cancels on both sides.

   1+cos θ = n²(1-cos θ)

3. Solve for cos θ. Collect the cos θ terms: cos θ (1+n²) = n²-1, so cos θ = (n²-1)/(n²+1).
4. So. θ = cos⁻¹((n²-1)/(n²+1)), which is option (d).

The idea to keep. For magnitudes of A⃗±B⃗, always work with squares: A² + B² ± 2AB cos θ. A ratio of magnitudes n becomes n² once squared.

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