Let $\vec{a}=\overset{^}{i}+2\overset{^}{j}+\overset{^}{k}$ and $\vec{b}=2\overset{^}{i}+\overset{^}{j}-\overset{^}{k}$. Let $\overset{^}{c}$ 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 $\overset{^}{c}$ is : [JEE Main 2025, 8 Apr (Shift 1)]

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YESPYQ · Accepted Answer

Correct answer: d) $\frac{1}{\sqrt{2}}(-\overset{^}{\mathrm{i}}+\overset{^}{\mathrm{k}})$. Sol: Let $\overset{‾}{c}=x\overset{‾}{a}+y\overset{‾}{b}$ $\begin{matrix}∣\overset{‾}{a}∣=\sqrt{6} \ ∣\overset{‾}{b}∣=\sqrt{6} \ \overset{‾}{a}\mathrm{.}\overset{‾}{b}=3\end{matrix}$ Apply let product with $\overset{‾}{a}$ $\begin{matrix}\overset{‾}{a}\mathrm{.}\overset{‾}{c}=x{\overset{‾}{a}}^{2}+y(\overset{‾}{a}\mathrm{.}\overset{‾}{b}) \ 0=6x+3y\Rightarrow y=−2x\ \ldots \ldots 1 \ {\overset{‾}{c}}^{2}={x}^{2}{\overset{‾}{a}}^{2}+{y}^{2}{\overset{‾}{b}}^{2}+2xt(\overset{‾}{a}\mathrm{.}\overset{‾}{b}) \ 1=6{x}^{2}+6{y}^{2}+6xy\ \ldots \ldots 2\end{matrix}$ By solving 1 & 2 $\begin{matrix}x=\pm \frac{1}{3\sqrt{2}}\ y=∓\frac{2}{3\sqrt{2}} \ ∴\overset{‾}{c}=\frac{1}{\sqrt{2}}(−\overset{^}{i}+\overset{^}{k})\end{matrix}$

Let \(\vec{a}=\overset{^}{i}+2\overset{^}{j}+\overset{^}{k}\) and \(\vec{b}=2\overset{^}{i}+\overset{^}{j}-\overset{^}{k…

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