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

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- $\sqrt{6}$

$\sqrt{6}$

$\sqrt{3}$

- $\sqrt{3}$

answer is C.

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Detailed Solution

$\begin{array}{l} Let \vec{v} = \hat{i} + \hat{j} + \hat{k} \\ \vec{b} \times \vec{c} = |\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 2 & 3 \\ 2 & 3 & - 1 \end{array}| \\ = - 7 \hat{i} + 7 \hat{j} + 7 \hat{k} \\ = - 7 (\hat{i} - \hat{j} - \hat{k}) \end{array}$
$Now \hat{a} = \frac{\hat{i} - \hat{j} - \hat{k}}{\sqrt{3}} or \frac{- \hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$
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$\begin{array}{l} \cos θ = \frac{\hat{a} \cdot \vec{v}}{| \vec{v} |} = \frac{1 - 1 - 1}{\sqrt{3} \sqrt{3}} = \frac{- 1}{3} \cos θ = \frac{\hat{a} \cdot \vec{v}}{| \vec{v} |} = \frac{- 1 + 1 + 1}{3} = \frac{1}{3} \\ (rejected) \\ \Rightarrow \hat{a} = \frac{\hat{i} - \hat{j} - \hat{k}}{\sqrt{3}} \end{array}$
$\begin{array}{l} Now \cos \frac{π}{3} = \frac{\hat{a} \cdot (\hat{i} + α \hat{j} + \hat{k})}{\sqrt{1 + α^{2} + 1}} \\ \Rightarrow \frac{1}{2} = \frac{1 - α - 1}{\sqrt{3} \sqrt{α^{2} + 2}} \\ \Rightarrow \frac{\sqrt{3}}{2} \sqrt{α^{2} + 2} = - α (∴ α < 0) \\ 3 α^{2} + 6 = 4 α^{2} \\ \Rightarrow α = - \sqrt{6} \end{array}$