JEE Main 2024 — Vector Algebra Question with Solution

Given: $\overrightarrow{a}=\hat{i}+\hat{j}+k$, $\overrightarrow{b}=-\hat{i}-8\hat{j}+2\hat{k}$ and $\overrightarrow{c}=4\hat{i}+c_2\hat{j}+c_{3}k$

Also, $\overrightarrow{b}\times \overrightarrow{a}=\overrightarrow{c}\times \overrightarrow{a}$

$\Rightarrow \left(\overrightarrow{b}-\overrightarrow{c}\right)\times \overrightarrow{a}=0$

$\Rightarrow \overrightarrow{b}-\overrightarrow{c}=\lambda \overrightarrow{a}$

$\Rightarrow \overrightarrow{b}=\overrightarrow{c}+\lambda \overrightarrow{a}$

$\Rightarrow -\hat{i}-8\hat{j}+2k=\left(4\hat{i}+c_2\hat{j}+c_{3}k\right)+\lambda \left(\hat{i}+\hat{j}+k\right)$

$\Rightarrow \lambda +4=-1, \lambda +c_2=-8, \lambda +c_3=2$

$\Rightarrow \lambda =-5, c_2=-3, c_3=7$

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

Now, finding angle we get,

$\cos \theta =\frac{\left(3\hat{i}+4\hat{j}+\hat{k}\right)·\left(4\hat{i}-3\hat{j}+7\hat{k}\right)}{\left|\left(3\hat{i}+4\hat{j}+\hat{k}\right)\right|\cdot \left|\left(4\hat{i}-3\hat{j}+7\hat{k}\right)\right|}$

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

$\Rightarrow \cos \theta =\frac{7}{\sqrt{26}\cdot \sqrt{74}}$

$\Rightarrow \cos \theta =\frac{7}{2\sqrt{481}}$

$\Rightarrow \sec \theta =\frac{2\sqrt{481}}{7}$

$\Rightarrow {\sec }^2\theta =\frac{4\times 481}{49}$

$\Rightarrow 1+{\tan }^2\theta =\frac{1924}{49}$

$\Rightarrow {\tan }^2\theta =\frac{1875}{49}$

$\Rightarrow {\tan }^2\theta \approx 38.26$

$\Rightarrow \left[{\tan }^2\theta \right]=\left[38.26\right]=38$