JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Three Dimensional Geometry

Points on the lines: $M=(4\lambda +5,\lambda +4,3\lambda +5)$ on $L_1$ (direction ${\vec{b}}_1=(4,1,3)$), and $N=(12\mu -8,5\mu -2,9\mu -11)$ on $L_2$ (direction ${\vec{b}}_2=(12,5,9)$).

For the shortest distance, $\overrightarrow{MN}$ is parallel to ${\vec{b}}_1\times {\vec{b}}_2$:

${\vec{b}}_1\times {\vec{b}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 4& 1& 3\\ 12& 5& 9\end{array}\right|=(9-15)\hat{i}-(36-36)\hat{j}+(20-12)\hat{k}=(-6,0,8)$.

$\overrightarrow{MN}=(12\mu -4\lambda -13,5\mu -\lambda -6,9\mu -3\lambda -16)$ must be proportional to $(-6,0,8)$.

Middle component $=0$: $5\mu -\lambda -6=0\Rightarrow \lambda =5\mu -6$.

Equating the ratio of first and third components: $\frac{12\mu -4\lambda -13}{-6}=\frac{9\mu -3\lambda -16}{8}$

$\Rightarrow 8(12\mu -4\lambda -13)=-6(9\mu -3\lambda -16)\Rightarrow 96\mu -32\lambda -104=-54\mu +18\lambda +96$

$\Rightarrow 150\mu -50\lambda =200\Rightarrow 3\mu -\lambda =4$.

With $\lambda =5\mu -6$: $3\mu -(5\mu -6)=4\Rightarrow -2\mu =-2\Rightarrow \mu =1,\lambda =-1$.

$M=(1,3,2)$, $N=(4,3,-2)$.

$\overrightarrow{OM}\cdot \overrightarrow{ON}=1(4)+3(3)+2(-2)=4+9-4=9$.