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JEE MAIN - Mathematics (2025 - 8th April Evening Shift - No. 18)

Let $ \vec{a} = \hat{i} + 2\hat{j} + \hat{k} $ and $ \vec{b} = 2\hat{i} + \hat{j} - \hat{k} $. Let $ \hat{c} $ be a unit vector in the plane of the vectors $ \vec{a} $ and $ \vec{b} $ and be perpendicular to $ \vec{a} $. Then such a vector $ \hat{c} $ is:
$ \frac{1}{\sqrt{2}}(-\hat{i} + \hat{k}) $
$ \frac{1}{\sqrt{5}}(\hat{j} - 2\hat{k}) $
$ \frac{1}{\sqrt{3}}(\hat{i} - \hat{j} + \hat{k}) $
$ \frac{1}{\sqrt{3}}(-\hat{i} + \hat{j} - \hat{k}) $

Explanation

Let vector $\vec{p}$ in plane of $\vec{a} \& \vec{b}=K(\vec{a}+\lambda \vec{b})$

$$\begin{aligned} & \overrightarrow{\mathrm{p}} \perp \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{a}}=0 \\ & \Rightarrow \mathrm{~K}(\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}}) \cdot \overrightarrow{\mathrm{a}}=0 \\ & \Rightarrow \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{a}}=0 \\ & \Rightarrow 6+\lambda(3)=0 \\ & \Rightarrow \lambda=-2 \\ & \Rightarrow \overrightarrow{\mathrm{p}}=(-3 \hat{\mathrm{i}}+3 \hat{\mathrm{k}}) \end{aligned}$$

Unit vector $\rightarrow \pm \frac{(-\hat{\mathrm{i}}+\hat{\mathrm{k}})}{\sqrt{2}}$

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