JEE Main 2024 — 1 February, Shift 2 — Question 25

$\quad \vec{a}=\hat{i}+\hat{j}+k$

$\overrightarrow{\mathrm{b}}=-\hat{\mathrm{i}}-8 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}$

$\overrightarrow{\mathrm{c}}=4 \hat{\mathrm{i}}+\mathrm{c}_{2} \hat{\mathrm{j}}+\mathrm{c}_{3} \mathrm{k}$

$\vec{b} \times \vec{a}=\vec{c} \times \vec{a}$

$(\vec{b}-\vec{c}) \times \vec{a}=0$

$\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}}=\lambda \vec{\alpha}$

$\overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{c}}+\lambda \vec{\alpha}$

$-\hat{\mathrm{i}}-8 \hat{\mathrm{j}}+2 \mathrm{k}=\left(4 \hat{\mathrm{i}}+\mathrm{c}_{2} \hat{\mathrm{j}}+\mathrm{c}_{3} \mathrm{k}\right)+\lambda(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\mathrm{k})$

$\lambda+4=-1 \Rightarrow \lambda=-5$

$\lambda+\mathrm{c}_{2}=-8 \Rightarrow \mathrm{c}_{2}=-3$

$\lambda+c_{3}=2 \Rightarrow c_{3}=7$

$\overrightarrow{\mathrm{c}}=4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+7 \mathrm{k}$

$\cos \theta=\frac{12-12+7}{\sqrt{26} \cdot \sqrt{74}}=\frac{7}{\sqrt{26} \cdot \sqrt{74}}=\frac{7}{2 \sqrt{481}}$

$\tan ^{2} \theta=\frac{625 \times 3}{49}$

$\left[\tan ^{2} \theta\right]=38$