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 :

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

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

(C) $\sin ^{-1}\left(\frac{n-1}{n+1}\right)$

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

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Vector

Two vectors

P

and

Q

have equal magnitudes. If the magnitude of

P + Q

n

times the magnitude of

P - Q

, then angle between

P

and

Q

is :

Answer: (B)

Let both vectors have magnitude

P

and let

θ

be the angle between them. Then

∣ P + Q ∣^{2} = 2 P^{2} (1 + cos θ), ∣ P - Q ∣^{2} = 2 P^{2} (1 - cos θ)

The condition

∣ P + Q ∣ = n ∣ P - Q ∣

squares to

1 + cos θ = n^{2} (1 - cos θ)

, so

cos θ = \frac{n ^{2} - 1}{n ^{2} + 1}

The angle between the two vectors is

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

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Single correct option