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 :

Correct answer: (B) cos^(-1)(n^2-1)/(-n^2-1). Given X=Y √(X^2+Y^2-2×Ycosθ) =n√(X^2+Y^2+2×Ycosθ) Square both sides 2X^2(1-cosθ)=n^2·2X^2(1+cosθ) 1-cosθ=n^2+n^2cosθ cosθ=(1-n^2)/(1+n^2) θ=cos^(-1)(n^2-1)/(-n^2-1)

Q80 / 90 · JEE Main 2021 (25 Jul Shift 2) · Physics

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 :

A
$\cos^{- 1} (\frac{- n^{2} - 1}{n^{2} - 1})$
Correct answer
B
$\cos^{- 1} (\frac{n^{2} - 1}{- n^{2} - 1})$
C
$\cos^{- 1} (\frac{n^{2} + 1}{- n^{2} - 1})$
D
$\cos^{- 1} (\frac{n^{2} + 1}{n^{2} - 1})$

Solution

Given $X = Y$

$\sqrt{X^{2} + Y^{2} - 2 \times Y \cos θ}$

$= n \sqrt{X^{2} + Y^{2} + 2 \times Y \cos θ}$

Square both sides

$2 X^{2} (1 - \cos θ) = n^{2} \cdot 2 X^{2} (1 + \cos θ)$

$1 - \cos θ = n^{2} + n^{2} \cos θ$

$\cos θ = \frac{1 - n^{2}}{1 + n^{2}}$

$θ = \cos^{- 1} [\frac{n^{2} - 1}{- n^{2} - 1}]$

Hence, the correct answer is option: 'B'.

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 :

Solution

More Mathematics in Physics Questions

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 :