JEE 2023 — Mathematics PYQ

The given lines are :

$$\frac{x-4}{4}=\frac{y+2}{5}=\frac{z+3}{3}\text{ and }\frac{x-1}{3}=\frac{y-3}{4}=\frac{z-4}{2}$$

$$\text{So, }{\vec{b}}_1=4\hat{i}+5\hat{j}+3\hat{k}$$
$${\vec{b}}_2=3\hat{i}+4\hat{j}+2\hat{k}$$
$${\vec{a}}_1=4\hat{i}-2\hat{j}-3\hat{k},{\vec{a}}_2=\hat{i}+3\hat{j}+4\hat{k}$$

$$\therefore {\vec{b}}_1\times {\vec{b}}_2=\left|\begin{array}{ccc}\hat{i}& amp;\hat{j}& amp;\hat{k}\\ 4& amp;5& amp;3\\ 3& amp;4& amp;2\end{array}\right|$$

$$=(10-12)\hat{i}-(8-9)\hat{j}+(16-15)\hat{k}$$
$$=-2\hat{i}+\hat{j}+\hat{k}$$

$$\text{Shortest distance, }d=\left|\frac{({\vec{a}}_2-{\vec{a}}_1)\cdot ({\vec{b}}_1\times {\vec{b}}_2)}{|{\vec{b}}_1\times {\vec{b}}_2|}\right|$$

$$=\left|\frac{(3\hat{i}-5\hat{j}-7\hat{k})\cdot (-2\hat{i}+\hat{j}+\hat{k})}{\sqrt{4+1+1}}\right|$$

$$=\left|\frac{-6-5-7}{\sqrt{6}}\right|=\frac{18}{\sqrt{6}}=3\sqrt{6}\text{ units}$$