JEE Main 2024 — Three Dimensional Geometry Question with Solution

Let,

$L_1:\frac{x+1}{1}=\frac{y}{{\displaystyle \frac{1}{2}}}=\frac{z}{{\displaystyle \frac{-1}{12}}},$

And $L_2:\frac{x}{1}=\frac{y+2}{1}=\frac{z-1}{\frac{1}{6}}$

Now, given $d_1=$ shortest distance between $L_1$ and $L_2$

Now using the formula of shortest distance we get,

 $d_1=\left|\frac{\left({\overrightarrow{a}}_2-{\overrightarrow{a}}_1\right)·\left({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2\right)}{\left|\left({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2\right)\right|}\right|$

$\Rightarrow d_1=\left|\frac{\left(\hat{i}-2\hat{j}+\hat{k}\right)·\left(\left(\hat{i}+\frac{1}{2}\hat{j}-\frac{1}{12}\hat{k}\right)\times \left(\hat{i}+\hat{j}+\frac{1}{6}\hat{k}\right)\right)}{\left|\left(\hat{i}+\frac{1}{2}\hat{j}-\frac{1}{12}\hat{k}\right)\times \left(\hat{i}+\hat{j}+\frac{1}{6}\hat{k}\right)\right|}\right|$

$\Rightarrow d_1=\left|\frac{\left(\hat{i}-2\hat{j}+\hat{k}\right)·\left(\frac{\hat{i}}{6}-\frac{\hat{j}}{4}+\frac{\hat{k}}{2}\right)}{\left|\left(\frac{\hat{i}}{6}-\frac{\hat{j}}{4}+\frac{\hat{k}}{2}\right)\right|}\right|$

$\Rightarrow d_1=\left|\frac{\frac{1}{6}+\frac{1}{2}+\frac{1}{2}}{\sqrt{\frac{49}{144}}}\right|$

$\Rightarrow d_1=\left|\frac{\frac{7}{6}}{\frac{7}{12}}\right|=2$

Now, let 

$L_3:\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5}, L_4:\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}$

Similarly finding $d_2=$ shortest distance between $L_3$ & $L_4$ we get,

$d_2=\left|\frac{\left(10\hat{j}+2\hat{k}\right)·\left(\left(2\hat{i}-7\hat{j}+5\hat{k}\right)\times \left(2\hat{i}+\hat{j}-3\hat{k}\right)\right)}{\left|\left(\left(2\hat{i}-7\hat{j}+5\hat{k}\right)\times \left(2\hat{i}+\hat{j}-3\hat{k}\right)\right)\right|}\right|$

$\Rightarrow d_2=\left|\frac{\left(10\hat{j}+2\hat{k}\right)·\left(16\hat{i}+16\hat{j}+16\hat{k}\right)}{\left|\left(16\hat{i}+16\hat{j}+16\hat{k}\right)\right|}\right|$

$\Rightarrow d_2=\left|\frac{160+32}{16\sqrt{3}}\right|$

$\Rightarrow d_2=\left|\frac{12}{\sqrt{3}}\right|$

Hence, $\frac{32\sqrt{3} d_1}{ d_2}=\frac{32\sqrt{3}\times 2}{\frac{12}{\sqrt{3}}}=16$