JEE MAIN - Mathematics (2024 - 1st February Evening Shift - No. 28)
Let $\overrightarrow{\mathrm{a}}=\hat{i}+\hat{j}+\hat{k}, \overrightarrow{\mathrm{b}}=-\hat{i}-8 \hat{j}+2 \hat{k}$ and $\overrightarrow{\mathrm{c}}=4 \hat{i}+\mathrm{c}_2 \hat{j}+\mathrm{c}_3 \hat{k}$ be three vectors such that $\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}$. If the angle between the vector $\overrightarrow{\mathrm{c}}$ and the vector $3 \hat{i}+4 \hat{j}+\hat{k}$ is $\theta$, then the greatest integer less than or equal to $\tan ^2 \theta$ is _______________.
Answer
38
Explanation
$\begin{aligned} & \vec{a}=\hat{i}+\hat{j}+k \\\\ & \vec{b}=\hat{i}+8 \hat{j}+2 k \\\\ & \vec{c}=4 \hat{i}+c_2 \hat{j}+c_3 k \\\\ & \vec{b} \times \vec{a}=\vec{c} \times \vec{a} \\\\ & (\vec{b}-\vec{c}) \times \vec{a}=0 \\\\ & \vec{b}-\vec{c}=\lambda \vec{\alpha} \\\\ & \vec{b}=\vec{c}+\lambda \vec{\alpha}\end{aligned}$
$\begin{aligned} & -\hat{\mathrm{i}}-8 \hat{\mathrm{j}}+2 \mathrm{k}=\left(4 \hat{\mathrm{i}}+\mathrm{c}_2 \hat{\mathrm{j}}+\mathrm{c}_3 \mathrm{k}\right)+\lambda(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\mathrm{k}) \\\\ & \lambda+4=-1 \Rightarrow \lambda=-5 \\\\ & \lambda+\mathrm{c}_2=-8 \Rightarrow \mathrm{c}_2=-3 \\\\ & \lambda+\mathrm{c}_3=2 \Rightarrow \mathrm{c}_3=7 \\\\ & \overrightarrow{\mathrm{c}}=4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+7 \mathrm{k}\end{aligned}$
$\begin{aligned} & \cos \theta=\frac{12-12+7}{\sqrt{26} \cdot \sqrt{74}}=\frac{7}{\sqrt{26} \cdot \sqrt{74}}=\frac{7}{2 \sqrt{481}} \\\\ & \tan ^2 \theta=\frac{625 \times 3}{49} \\\\ & {\left[\tan ^2 \theta\right]=38}\end{aligned}$
$\begin{aligned} & -\hat{\mathrm{i}}-8 \hat{\mathrm{j}}+2 \mathrm{k}=\left(4 \hat{\mathrm{i}}+\mathrm{c}_2 \hat{\mathrm{j}}+\mathrm{c}_3 \mathrm{k}\right)+\lambda(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\mathrm{k}) \\\\ & \lambda+4=-1 \Rightarrow \lambda=-5 \\\\ & \lambda+\mathrm{c}_2=-8 \Rightarrow \mathrm{c}_2=-3 \\\\ & \lambda+\mathrm{c}_3=2 \Rightarrow \mathrm{c}_3=7 \\\\ & \overrightarrow{\mathrm{c}}=4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+7 \mathrm{k}\end{aligned}$
$\begin{aligned} & \cos \theta=\frac{12-12+7}{\sqrt{26} \cdot \sqrt{74}}=\frac{7}{\sqrt{26} \cdot \sqrt{74}}=\frac{7}{2 \sqrt{481}} \\\\ & \tan ^2 \theta=\frac{625 \times 3}{49} \\\\ & {\left[\tan ^2 \theta\right]=38}\end{aligned}$
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