Two vectors A and B have equal magnitudes. The magnitude of (A + B) is 'n' times the magnitude of (A - B). The angle between A and B is - (a) sin ⁻¹[(n-1)/(n+1)] (b) sin ⁻¹[(n²-1)/(n²+1)] (c) cos ⁻¹[(n²-1)/(n²+1)] (d) cos ⁻¹[(n-1)/(n+1)]

Answer: (c). |A + B| = 2Acos (θ)/(2), |A - B| = 2Asin (θ)/(2), so cot (θ)/(2) = n. cos θ = (n² - 1)/(n² + 1) So the answer is (c) cos ⁻¹[(n² - 1)/(n² + 1)].

Two vectors A and B have equal magnitudes. The magnitude of (A + B)… | JEE Main 2019 (10 Jan, Shift 2)

Q29JEE Main 2019 · 10 JAN · SHIFT 2Medium

Addition & Subtraction of Vectors

Two vectors $\vec{A}$ and $\vec{B}$ have equal magnitudes. The magnitude of $\left(\vec{A} + \vec{B}\right)$ is 'n' times the magnitude of $\left(\vec{A} - \vec{B}\right)$ . The angle between $\vec{A}$ and $\vec{B}$ is -

Show answer and solution

Answer: (c)

$|\vec{A} + \vec{B}| = 2A\cos\frac{\theta}{2}$ , $|\vec{A} - \vec{B}| = 2A\sin\frac{\theta}{2}$ , so $\cot\frac{\theta}{2} = n$ .
$\cos\theta = \frac{n^2 - 1}{n^2 + 1}$
So the answer is (c) $\cos^{-1}\left[\frac{n^2 - 1}{n^2 + 1}\right]$ .

Report a mistake

Addition & Subtraction of Vectors question from JEE Main 2019 (10 Jan, Shift 2)

More questions on Addition & Subtraction of Vectors