Two vectors P and Q have equal magnitudes. If the magnitude of P + Q is n times the magnitude of P - Q, then angle between P and Q is: (a) sin ⁻¹((n-1)/(n+1)) (b) cos ⁻¹((n-1)/(n+1)) (c) sin ⁻¹((n²-1)/(n²+1)) (d) cos ⁻¹((n²-1)/(n²+1))

Answer: (d). Equal magnitudes: (|P + Q|)/(|P - Q|) = cot (θ)/(2) = n cos θ = (cot ²(θ/2) - 1)/(cot ²(θ/2) + 1) = (n² - 1)/(n² + 1) So the answer is (d) cos ⁻¹((n² - 1)/(n² + 1)).

Two vectors P and Q have equal magnitudes. If the magnitude of P + Q… | JEE Main 2021 (20 Jul, Shift 2)

Q19JEE Main 2021 · 20 JUL · SHIFT 2Medium

Addition & Subtraction of Vectors

Two vectors $\vec{P}$ and $\vec{Q}$ have equal magnitudes. If the magnitude of $\vec{P} + \vec{Q}$ is n times the magnitude of $\vec{P} - \vec{Q}$ , then angle between $\vec{P}$ and $\vec{Q}$ is:

Show answer and solution

Answer: (d)

Equal magnitudes: $\frac{|\vec{P} + \vec{Q}|}{|\vec{P} - \vec{Q}|} = \cot\frac{\theta}{2} = n$
$\cos\theta = \frac{\cot^2(\theta/2) - 1}{\cot^2(\theta/2) + 1} = \frac{n^2 - 1}{n^2 + 1}$
So the answer is (d) $\cos^{-1}\left(\frac{n^2 - 1}{n^2 + 1}\right)$ .

Report a mistake

Addition & Subtraction of Vectors question from JEE Main 2021 (20 Jul, Shift 2)

More questions on Addition & Subtraction of Vectors