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 :

(C) $\cos^{-1}\left(\frac{n^{2}-1}{-n^{2}-1}\right)$

(A) $\cos^{-1}\left(\frac{n^{2}+1}{n^{2}-1}\right)$

(B) $\cos^{-1}\left(\frac{-n^{2}-1}{n^{2}-1}\right)$

(D) $\cos^{-1}\left(\frac{n^{2}+1}{-n^{2}-1}\right)$

JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Vector

Two vectors

X

and

Y

have equal magnitude. The magnitude of

(X - Y)

n

times the magnitude of

(X + Y)

. The angle between

X

and

Y

is :

Answer: (C)

Writing

∣ X ∣ = ∣ Y ∣ = a

and

θ

for the angle between them,

∣ X - Y ∣^{2} = 2 a^{2} (1 - cos θ), ∣ X + Y ∣^{2} = 2 a^{2} (1 + cos θ)

The condition

∣ X - Y ∣ = n ∣ X + Y ∣

gives

1 - cos θ = n^{2} (1 + cos θ)

, so

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

The angle between the vectors is

cos^{- 1} (\frac{n ^{2} - 1}{- n ^{2} - 1})

JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Single correct option