JEE Main 2024 — 8 April, Shift 2 — Question 8

$\begin{array}{c}\overline{\mathrm{r}}_{1}=(\lambda \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\alpha(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})\end{array}$

$\overline{\mathrm{r}}_{2}=(2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+7 \hat{\mathrm{k}})+\beta(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})$

$\begin{aligned} & \overline{\mathrm{b}}=2 \hat{\mathrm{i}}+3 \mathrm{j}+4 \hat{\mathrm{k}} &\end{aligned}$

$\bar{a}_{1}+\lambda \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$

$\overline{\mathrm{a}}_{2}=2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}$

Shortest dist. $=\frac{\left|\overline{\mathrm{b}} \times\left(\overline{\mathrm{a}}_{2}-\overline{\mathrm{a}}_{1}\right)\right|}{|\mathrm{b}|}=\frac{13}{\sqrt{29}}$

$\frac{|(2 \hat{i}+3 \hat{j}+4 \hat{k}) \times((2-\lambda) \hat{i}+4 \hat{k})|}{\sqrt{29}}=\frac{13}{\sqrt{29}}$

$|-8 \hat{\mathrm{j}}-3(2-\lambda) \hat{\mathrm{k}}+12 \hat{\mathrm{i}}+4(2-\lambda) \hat{\mathrm{j}}|=13$

$|12 \hat{i}-4 \lambda \hat{j}+(3 \lambda-6) \hat{k}|=13$

$144+16 \lambda^{2}+(3 \lambda-6)^{2}=169$

$16 \lambda^{2}+(3 \lambda-6)^{2}=25$ $\lambda =1$