JEE Main 2024 — 27 January, Shift 2 — Question 3

$\mathrm{L}_{1}=\frac{\mathrm{x}}{1}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}-2}{3}=\lambda$

$\mathrm{M}(\lambda, 1+2 \lambda, 2+3 \lambda)$

$\overrightarrow{\mathrm{PM}}=(\lambda-1) \hat{\mathrm{i}}+(1+2 \lambda) \hat{\mathrm{j}}+(3 \lambda-5) \hat{\mathrm{k}}$ $\overrightarrow{\mathrm{PM}}$

is perpendicular to line $\mathrm{L}_{1}$

$\overrightarrow{\mathrm{PM}} \cdot \overrightarrow{\mathrm{b}}=0 \quad(\overrightarrow{\mathrm{~b}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})$

$\Rightarrow \lambda-1+4 \lambda+2+9 \lambda-15=0$

$14 \lambda=14 \Rightarrow \lambda=1$

$\therefore \mathrm{M}=(1,3,5)$

$\overrightarrow{\mathrm{Q}}=2 \overrightarrow{\mathrm{M}}-\overrightarrow{\mathrm{P}}[\mathrm{M}$

is midpoint of $\overrightarrow{\mathrm{P}} \& \overrightarrow{\mathrm{Q}}]$

$\overrightarrow{\mathrm{Q}}=2 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+10 \hat{\mathrm{k}}-\hat{\mathrm{i}}-7 \hat{\mathrm{k}}$

$\overrightarrow{\mathrm{Q}}=\hat{\mathrm{i}}+6 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$

$\therefore(\alpha, \beta, \gamma)=(1,6,3)$

Required line having direction cosine $(l, \mathrm{~m}, \mathrm{n})$

$l^{2}+m^{2}+n^{2}=1$ $\Rightarrow l^{2}+\left(-\frac{1}{2}\right)^{2}+\left(-\frac{1}{\sqrt{2}}\right)^{2}=1$ $l^{2}=\frac{1}{4}$ $\therefore l=\frac{1}{2}$

Line make acute angle with x -axis] Equation of line passing through $(1,6,3)$ \

will be $\overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}+6 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\mu\left(\frac{1}{2} \hat{\mathrm{i}}-\frac{1}{2} \hat{\mathrm{j}}-\frac{1}{\sqrt{2}} \hat{\mathrm{k}}\right)$ Option (3) satisfying for $\mu=4$