Let $\vec{a}=\hat{i}+2 \hat{j}+\hat{k}, \vec{b}=3 \hat{i}-3 \hat{j}+3 \hat{k}$,…

Let $\vec{a}=\hat{i}+2 \hat{j}+\hat{k}, \vec{b}=3 \hat{i}-3 \hat{j}+3 \hat{k}$, $\overrightarrow{\mathrm{c}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{d}}$ be a vector such that $\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{d}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{d}}$ and $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{d}}=4$. Then $|(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{d}})|^2$ is equal to ______ .

Solution

$\begin{aligned} & \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{d}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{d}} \text {-and } \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{d}}=4 \\ & \Rightarrow \overrightarrow{\mathrm{~d}}=\lambda(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}})=\lambda(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}) \\ & \because \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{d}}=4 \Rightarrow \lambda=-2 \\ & \text { Also. }|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{d}}|^2+|\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{d}}|^2=|\overrightarrow{\mathrm{a}}|^2|\overrightarrow{\mathrm{~d}}|^2 \\ & \Rightarrow|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{d}}|^2=6 \times 4 \times 6-16=128\end{aligned}$

Asked in: JEE Main 2025 (03 Apr Shift 2)

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