Two vectors X and Y have equal magnitude. The magnitude of ( X - Y) is n times the magnitude of ( X + Y). The angle between X and Y is: (a) cos ⁻¹((-n² - 1)/(n² - 1)) (b) cos ⁻¹((n² - 1)/(-n² - 1)) (c) cos ⁻¹((n² + 1)/(-n² - 1)) (d) cos ⁻¹((n² + 1)/(n² - 1))

Answer: (b). Equal magnitudes: (|X - Y|)/(|X + Y|) = tan (θ)/(2) = n cos θ = (1 - n²)/(1 + n²) = (n² - 1)/(-n² - 1) So the answer is (b).

Two vectors X and Y have equal magnitude. The magnitude of ( X - Y)… | JEE Main 2021 (25 Jul, Shift 2)

Q18JEE Main 2021 · 25 JUL · SHIFT 2Medium

Addition & Subtraction of Vectors

Two vectors $\vec{X}$ and $\vec{Y}$ have equal magnitude. The magnitude of ( $\vec{X} - \vec{Y}$ ) is n times the magnitude of ( $\vec{X} + \vec{Y}$ ). The angle between $\vec{X}$ and $\vec{Y}$ is:

Show answer and solution

Answer: (b)

Equal magnitudes: $\frac{|\vec{X} - \vec{Y}|}{|\vec{X} + \vec{Y}|} = \tan\frac{\theta}{2} = n$
$\cos\theta = \frac{1 - n^2}{1 + n^2} = \frac{n^2 - 1}{-n^2 - 1}$
So the answer is (b).

Report a mistake

Addition & Subtraction of Vectors question from JEE Main 2021 (25 Jul, Shift 2)

More questions on Addition & Subtraction of Vectors