Let â be a unit vector perpendicular to the vectors $\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{c}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}}$, and makes an angle of $\cos ^{-1}\left(-\frac{1}{3}\right)$ with the vector $\hat{i}+\hat{j}+\hat{k}$. If $\hat{a}$ makes an angle of $\frac{\pi}{3}$ with the vector $\hat{\mathrm{i}}+\alpha \hat{\mathrm{j}}+\hat{\mathrm{k}}$, then the value of $\alpha$ is :

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Vector Algebra

Let â be a unit vector perpendicular to the vectors

b = \hat{i} - 2 \hat{j} + 3 \hat{k}

and

c = 2 \hat{i} + 3 \hat{j} - \hat{k}

, and makes an angle of

cos^{- 1} (- \frac{1}{3})

with the vector

\hat{i} + \hat{j} + \hat{k}

. If

\overset{a}{^}

makes an angle of

\frac{π}{3}

with the vector

\hat{i} + α \hat{j} + \hat{k}

, then the value of

α

is :

Answer: (C)

b \times c = \hat{i} 12 \hat{j} - 2 3 \hat{k} 3 - 1 = (2 - 9, - (- 1 - 6), 3 + 4) = (- 7, 7, 7)

\overset{a}{^} = \pm \frac{( 1 , - 1 , - 1 )}{3}

. Checking against

v = \hat{i} + \hat{j} + \hat{k}

:

for

\overset{a}{^} = \frac{( 1 , - 1 , - 1 )}{3}

cos θ = \frac{1 - 1 - 1}{3 \cdot 3} = - \frac{1}{3}

, matching the given angle. So

\overset{a}{^} = \frac{i ^ - j ^ - k ^}{3}

.

For the angle with

\hat{i} + α \hat{j} + \hat{k}

cos \frac{π}{3} = \frac{1 - α - 1}{3 α ^{2} + 2} = \frac{- α}{3 α ^{2} + 2} = \frac{1}{2}

Since the left side must be positive,

α < 0

, and squaring:

\frac{3 ( α ^{2} + 2 )}{4} = α^{2} \Rightarrow 3 α^{2} + 6 = 4 α^{2} \Rightarrow α^{2} = 6 \Rightarrow α = - 6

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Single correct option