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JEE MAIN 2024
1-2-2024 S2
Question
Let $\vec{a}=\hat{\imath}+\hat{\jmath}+\hat{k}, \vec{b}=-\hat{\imath}-8 \hat{\jmath}+2 \hat{k}$ and $\overrightarrow{\mathrm{c}}=4 \hat{\imath}+\mathrm{c}_2 \hat{\jmath}+\mathrm{c}_3 \hat{\mathrm{k}}$ be three vectors such that $\vec{b} \times \vec{a}=\vec{c} \times \vec{a}$. If the angle between the vector $\overrightarrow{\mathrm{c}}$ and the vector $3 \hat{\mathrm{i}}+4 \hat{\mathbf{j}}+\hat{\mathrm{k}}$ is $\theta$, then the greatest integer less than or equal to $\tan ^2 \theta$ is :
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Solution
$\begin{aligned} & \vec{a}=\hat{\imath}+\hat{\jmath}+k \\ & \overrightarrow{\mathrm{~b}}=\hat{\imath}+8 \hat{\jmath}+2 \mathrm{k} \\ & \overrightarrow{\mathrm{c}}=4 \hat{\imath}+\mathrm{c}_2 \hat{\jmath}+\mathrm{c}_3 \mathrm{k} \\ & \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}} \\ & (\vec{b}-\vec{c}) \times \vec{a}=0 \\ & \overrightarrow{\mathrm{~b}}-\overrightarrow{\mathrm{c}}=\lambda \vec{\alpha} \\ & \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{c}}+\lambda \vec{\alpha} \\ & -\hat{\imath}-8 \hat{\jmath}+2 k=\left(4 \hat{\imath}+c_2 \hat{\jmath}+c_3 k\right)+\lambda(\hat{\imath}+\hat{\jmath}+k) \\ & \lambda+4=-1 \Rightarrow \lambda=-5 \\ & \lambda+\mathrm{c}_2=-8 \Rightarrow \mathrm{c}_2=-3 \\ & \lambda+\mathrm{c}_3=2 \Rightarrow \mathrm{c}_3=7 \\ & \overrightarrow{\mathrm{c}}=4 \hat{\imath}-3 \hat{\mathrm{j}}+7 \mathrm{k} \\ & \cos \theta=\frac{12-12+7}{\sqrt{26} \cdot \sqrt{74}}=\frac{7}{\sqrt{26} \cdot \sqrt{74}}=\frac{7}{2 \sqrt{481}} \\ & \tan ^2 \theta=\frac{625 \times 3}{49} \\ & {\left[\tan ^2 \theta\right]=38}\end{aligned}$
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