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

Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=3 \hat{i}+2 \hat{j}-\hat{k}, \vec{c}=\lambda \hat{j}+\mu \hat{k}$ and $\hat{d}$ be a unit vector such that $\overrightarrow{\mathrm{a}} \times \hat{\mathrm{d}}=\overrightarrow{\mathrm{b}} \times \hat{\mathrm{d}}$ and $\overrightarrow{\mathrm{c}} \cdot \hat{\mathrm{d}}=1$, If $\vec{c}$ is perpendicular to $\vec{a}$, then $|3 \lambda \hat{d}+\mu \overrightarrow{\mathrm{c}}|^2$ is equal to _______ .

Solution

$\begin{aligned} & \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{d}}-\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{d}}=0 \\ & (\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{d}}=0 \\ & \overrightarrow{\mathrm{~d}}=\mathrm{t}(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}) \\ & \overrightarrow{\mathrm{d}}=\mathrm{t}(-2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \\ & |\overrightarrow{\mathrm{d}}|=1 \\ & |\mathrm{t}|=\frac{1}{3}\end{aligned}$
$\begin{aligned} & \overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}=0 \\ & \lambda+\mu=0 \\ & \mu=-\lambda \\ & \overrightarrow{\mathrm{c}}=\lambda(\hat{\mathrm{j}}-\hat{\mathrm{k}}),|\overrightarrow{\mathrm{c}}|^2=2 \lambda^2 \\ & \overrightarrow{\mathrm{c}} \cdot \hat{\mathrm{d}}=1 \\ & \mathrm{t}(-2,-1,2) \cdot \lambda(0,1,-1)=1 \\ & \lambda \mathrm{t}=\frac{-1}{3} \Rightarrow \lambda^2=1\end{aligned}$
$\begin{aligned} & |3 \lambda \hat{\mathrm{~d}}+\mu \mathrm{c}|^2=9 \lambda^2|\hat{\mathrm{~d}}|^2+\mu^2|\overrightarrow{\mathrm{c}}|^2+6 \lambda \mu(\hat{\mathrm{~d}} \cdot \overrightarrow{\mathrm{c}}) \\ & =3 \lambda^2+2 \lambda^4 \\ & =5\end{aligned}$

Asked in: JEE Main 2025 (03 Apr Shift 1)

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