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Let $\vec{a}=2 \hat{i}+\hat{j}-2 \hat{k}, \vec{b}=\hat{i}+\hat{j}$ and $\vec{c}=\vec{a} \times \vec{b}$. Let $\vec{d}$ be a vector such that $|\overrightarrow{\mathrm{d}}-\overrightarrow{\mathrm{a}}|=\sqrt{11},|\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{d}}|=3$ and the angle between $\overrightarrow{\mathrm{c}}$ and $\overrightarrow{\mathrm{d}}$ is $\frac{\pi}{4}$. Then $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{d}}$ is equal to
[JEE MAIN 24 Jan 2026 S I]
(1) 11
(2) 3
(3) 0
(4) 1Solution
\[ \begin{aligned} & \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{array}\right|=2 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}} \\ & |\overrightarrow{\mathrm{c}}|=\sqrt{4+4+1}=3 \\ & |\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{d}}|=|\overrightarrow{\mathrm{c}} \| \overrightarrow{\mathrm{d}}| \sin \frac{\pi}{4}=3 \\ & \Rightarrow 3|\overrightarrow{\mathrm{~d}}| \frac{1}{\sqrt{2}}=3 \Rightarrow|\overrightarrow{\mathrm{~d}}|=\sqrt{2} \\ & \operatorname{given}|\overrightarrow{\mathrm{~d}}-\overrightarrow{\mathrm{a}}|=\sqrt{11} \Rightarrow(\overrightarrow{\mathrm{~d}}-\overrightarrow{\mathrm{a}})^{2}=11 \\ & \Rightarrow|\overrightarrow{\mathrm{~d}}|^{2}+|\overrightarrow{\mathrm{a}}|^{2}-2 \overrightarrow{\mathrm{~d}} \cdot \overrightarrow{\mathrm{a}}=11 \\ & \Rightarrow 2+(4+1+4)-2 \overrightarrow{\mathrm{~d}} \cdot \overrightarrow{\mathrm{a}}=11 \\ & \Rightarrow \overrightarrow{\mathrm{~d}} \cdot \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{~d}}=0 \text { Answer }(3) \end{aligned} \]
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