QJEEM24Ma3099: JEE Main 2024 Mathematics - Class 12 JEEM24Ma3099 Solution: Let O be the origin, and M and be the points on the lines and…

$\mathrm{L}_1: \frac{\mathrm{x}-5}{4}=\frac{\mathrm{y}-4}{1}=\frac{\mathrm{z}-5}{3}=\lambda \quad \operatorname{drs}(4,1,3)=\mathrm{b}_1$

$\begin{aligned}& \mathrm{M}(4 \lambda+5, \lambda+4,3 \lambda+5) \\& \mathrm{L}_2: \frac{\mathrm{x}+8}{12}=\frac{\mathrm{y}+2}{5}=\frac{\mathrm{z}+11}{9}=\mu \\& \mathrm{N}(12 \mu-8,5 \mu-2,9 \mu-11) \\& \overrightarrow{\mathrm{MN}}=(4 \lambda-12 \mu+13, \lambda-5 \mu+6,3 \lambda-9 \mu+16) \quad \text{..... (1)}\end{aligned}$

Now

$\overrightarrow{\mathrm{b}}_1 \times \overrightarrow{\mathrm{b}}_2=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\4 & 1 & 3 \\12 & 5 & 9\end{array}\right|=-6 \hat{\mathrm{i}}+8 \hat{\mathrm{k}} \quad \text{.... (2)}$

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

$\lambda-5 \mu+6=0 \quad \text{.... (3)}$

I and III

$\lambda-3 \mu+4=0 \quad \text{.... (4)}$

Solve (3) and (4) we get

$\begin{aligned}\lambda= & -1, \mu=1 \\\therefore \quad & \mathrm{M}(1,3,2) \\& \mathrm{N}(4,3,-2) \\\therefore \quad & \overrightarrow{\mathrm{OM}} \cdot \overrightarrow{\mathrm{ON}}=4+9-4=9\end{aligned}$