Two vectors $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$ have equal magnitudes. If magnitude of $\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}$ is equal to two times the magnitude of $\vec{A}-\vec{B}$, then the angle between $\vec{A}$ and $\vec{B}$ will be :

(C) $\cos ^{-1}\left(\frac{3}{5}\right)$

(A) $\sin ^{-1}\left(\frac{3}{5}\right)$

(B) $\sin ^{-1}\left(\frac{1}{3}\right)$

(D) $\cos ^{-1}\left(\frac{1}{3}\right)$

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Vector

Two vectors

A

and

B

have equal magnitudes. If magnitude of

A + B

is equal to two times the magnitude of

A - B

, then the angle between

A

and

B

will be :

Answer: (C)

For two vectors of equal magnitude

a

, we have

∣ A + B ∣ = 2∣ A - B ∣

. Using the magnitude formula:

a^{2} + a^{2} + 2 a^{2} cos θ = 2 a^{2} + a^{2} - 2 a^{2} cos θ

Squaring both sides:

2 a^{2} (1 + cos θ) = 4 \cdot 2 a^{2} (1 - cos θ)

1 + cos θ = 4 (1 - cos θ)

1 + cos θ = 4 - 4 cos θ

5 cos θ = 3

cos θ = \frac{3}{5}

Therefore,

θ = cos^{- 1} (\frac{3}{5})

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Single correct option