Consider two vectors $\overrightarrow{\mathrm{u}}=3 \hat{\mathrm{i}}-\hat{\mathrm{j}}$ and $\overrightarrow{\mathrm{v}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\lambda \hat{\mathrm{k}}, \lambda>0$. The angle between them is given by $\cos ^{-1}\left(\frac{\sqrt{5}}{2 \sqrt{7}}\right)$. Let $\vec{v}=\vec{v}_{1}+\vec{v}_{2}$, where $\vec{v}_{1}$ is parallel to $\overrightarrow{\mathrm{u}}$ and $\overrightarrow{\mathrm{v}}_{2}$ is perpendicular to $\overrightarrow{\mathrm{u}}$. Then the value $\left|\overrightarrow{\mathrm{v}}_{1}\right|^{2}+\left|\overrightarrow{\mathrm{v}}_{2}\right|^{2}$ is equal to

JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Vector Algebra

Consider two vectors

u = 3 \hat{i} - \hat{j}

and

v = 2 \hat{i} + \hat{j} - λ \hat{k}, λ > 0

. The angle between them is given by

cos^{- 1} (\frac{5}{2 7})

. Let

v = v_{1} + v_{2}

, where

v_{1}

is parallel to

u

and

v_{2}

is perpendicular to

u

. Then the value

v_{1}^{2} + v_{2}^{2}

is equal to

Answer: (B)

The angle satisfies

cos θ = \frac{u \cdot v}{∣ u ∣∣ v ∣} = \frac{5}{10 5 + λ ^{2}}

, and equating this to

\frac{5}{2 7}

and squaring gives

\frac{25}{10 ( 5 + λ ^{2} )} = \frac{5}{28}

, so

5 + λ^{2} = 14

and

λ = 3

. Then

∣ v ∣^{2} = 4 + 1 + 9 = 14

. Splitting

v = v_{1} + v_{2}

with

v_{1} ∥ u

and

v_{2} ⊥ u

makes

v_{1}

and

v_{2}

perpendicular to each other, so

∣ v_{1} ∣^{2} + ∣ v_{2} ∣^{2} = ∣ v ∣^{2} = 14

JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Single correct option