Question

If the shortest distance between the lines
$\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}$
and $\frac{x+3}{3}=\frac{y+7}{2}=\frac{z-6}{4}$
is $\lambda \sqrt{30}$ units, then the value of $\lambda$ is :

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (C)

$3$

Let there be two lines $\overrightarrow{r_1}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1},\overrightarrow{r_2}=\overrightarrow{a_2}+\lambda \overrightarrow{b_2}$
if $\overrightarrow{b_1},\overrightarrow{b_2}$ are not collinear
Then shortest distance between the line would $|\frac{(\overrightarrow{a_2}-\overrightarrow{a_1}).(\overrightarrow{b_1}\times \overrightarrow{b_2})}{|\overrightarrow{b_1}\times \overrightarrow{b_2}|}|$
Two lines$\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}=t$ and $\frac{x+3}{3}=\frac{y+7}{2}=\frac{z-6}{4}=t$
$⟹$ First line $x\hat{i}+y\hat{j}+z\hat{k}=3\hat{i}+8\hat{j}+3\hat{k}+t(3\hat{i}-\hat{j}+\hat{k})$
$⟹$ Second line $x\hat{i}+y\hat{j}+z\hat{k}=-3\hat{i}-7\hat{j}+6\hat{k}+t(3\hat{i}+2\hat{j}+4\hat{k})$
Hence we got $\overrightarrow{a_1}=3\hat{i}+8\hat{j}+3\hat{k},\overrightarrow{b_1}=3\hat{i}-\hat{j}+\hat{k}$
$\overrightarrow{a_2}=-3\hat{i}-7\hat{j}+6\hat{k},\overrightarrow{b_2}=3\hat{i}+2\hat{j}+4\hat{k}$
So shortest dist is $|\frac{(\overrightarrow{a_2}-\overrightarrow{a_1}).(\overrightarrow{b_1}\times \overrightarrow{b_2})}{|\overrightarrow{b_1}\times \overrightarrow{b_2}|}|$
Shortest dist=$|\frac{\left(\right(-3\hat{i}-7\hat{j}+6\hat{k})-(3\hat{i}+8\hat{j}+3\hat{k}\left)\right).\left(\right(3\hat{i}-\hat{j}+\hat{k})\times (3\hat{i}+2\hat{j}+4\hat{k}\left)\right)}{|(3\hat{i}-\hat{j}+\hat{k})\times (3\hat{i}+2\hat{j}+4\hat{k})|}|$
Shortest Dist=$|\frac{(-6\hat{i}-15\hat{j}+3\hat{k}).(-6\hat{i}+9\hat{j}+9\hat{k})}{|-6\hat{i}+9\hat{j}+9\hat{k}|}|=|\frac{-72}{\sqrt{208}}|=|\frac{18}{\sqrt{13}}|$