JEE Main 2024 — 29 January, Shift 2 — Question 26

$\quad \mathrm{L}_{1}: \frac{\mathrm{x}-5}{4}=\frac{\mathrm{y}-4}{1}=\frac{\mathrm{z}-5}{3}=\lambda \quad$

$\operatorname{drs}(4,1,3)=b_{1}$

$\mathrm{M}(4 \lambda+5, \lambda+4,3 \lambda+5)$

$L_{2}: \frac{x+8}{12}=\frac{y+2}{5}=\frac{z+11}{9}=\mu$

$\mathrm{N}(12 \mu-8,5 \mu-2,9 \mu-11)$

$\overrightarrow{\mathrm{MN}}=(4 \lambda-12 \mu+13, \lambda-5 \mu+6,3 \lambda-9 \mu+16)$

Now $\overrightarrow{\mathrm{b}}_{1} \times \overrightarrow{\mathrm{b}}_{2}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}}\\ 4 & 1 & 3 \\12 & 5 & 9\end{array}\right|=-6 \hat{\mathrm{i}}+8 \hat{\mathrm{k}}$

Equation (1) and (2) $\therefore \frac{4 \lambda-12 \mu+13}{-6}=\frac{\lambda-5 \mu+6}{0}=\frac{3 \lambda-9 \mu+16}{8}$

I and II

$\lambda-5 \mu+6=0$

I and III

$\lambda-3 \mu+4=0$

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

$$\mathrm{N}(4,3,-2)$$

$\therefore \overrightarrow{\mathrm{OM}} \cdot \overrightarrow{\mathrm{ON}}=4+9-4=9$