Let be a unit vector perpendicular to the vectors and , and makes an angle of with the vector . If makes an angle of with the vector , then the value of is : [2025]

Question

Let $\hat{a}$ 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 $\hat{a}$ makes an angle of $\frac{π}{3}$ with the vector $\hat{i} + α \hat{j} + \hat{k}$ , then the value of $α$ is : [2025]

Smart Achievers · Accepted Answer

(3)

Let $\vec{m} = \hat{i} + \hat{j} + \hat{k}$

Since, $\vec{b} \times \vec{c} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 2 & 3 \\ 2 & 3 & - 1 \end{matrix} |$ $= - 7 \hat{i} + 7 \hat{j} + 7 \hat{k} = - 7 (\hat{i} - \hat{j} - \hat{k})$

$∵ \hat{a} parallel to \hat{b} \times \hat{c}$

$∴ \hat{a} = \frac{\hat{i} - \hat{j} - \hat{k}}{\sqrt{3}} or \frac{- \hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$

Now, $\cos θ = \frac{\hat{a} \cdot \vec{m}}{| \vec{m} |} = \frac{1 - 1 - 1}{\sqrt{3} \sqrt{3}} = \frac{- 1}{3}$ ${∵ \cos^{- 1} (\frac{- 1}{3}) = θ}$

or $\cos θ = \frac{- 1 + 1 + 1}{\sqrt{3} \sqrt{3}} = \frac{1}{3}$ (rejected)

Hence, $\hat{a} = \frac{\hat{i} - \hat{j} - \hat{k}}{\sqrt{3}}$

Now, $\cos (\frac{π}{3}) = \frac{\hat{a} \cdot (\hat{i} + α \hat{j} + \hat{k})}{\sqrt{1 + α^{2} + 1}} \Rightarrow \frac{1}{2} = \frac{1 - α - 1}{\sqrt{3} \sqrt{α^{2} + 2}}$

$\Rightarrow \sqrt{3} \sqrt{α^{2} + 2} = - 2 α \Rightarrow α = \frac{- \sqrt{3 (α^{2} + 2)}}{2}, i . e ., α < 0$

$\Rightarrow 3 α^{2} + 6 = 4 α^{2} \Rightarrow α = - \sqrt{6}$ .

Solution

1 $\sqrt{3}$

2 $\sqrt{6}$

3 $- \sqrt{6}$

4 $- \sqrt{3}$

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Solution

Q. Let a^ be a unit vector perpendicular to the vectors b→=i^–2j^+3k^ and c→=2i^+3j^–k^, and makes an angle of cos–1(–13) with the vector i^+j^+k^. If a^ makes an angle of π3 with the vector i^+αj^+k^, then the value of α is : [2025]

1 3

2 6

3 –6

4 –3

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1 $\sqrt{3}$

2 $\sqrt{6}$

3 $- \sqrt{6}$

4 $- \sqrt{3}$