Question

The shortest distance between the skew lines $\overrightarrow{r}=3\hat{i}+8\hat{j}+3\hat{k}+\alpha (3\hat{i}-\hat{j}+\hat{k})$ and $\overrightarrow{r}=-3\hat{i}-7\hat{j}+6\hat{k}+\beta (-3\hat{i}+2\hat{j}+4\hat{k})$ is

• A. $\sqrt{30}$
• B. $2\sqrt{30}$
• C. $3\sqrt{30}$
• D. $4\sqrt{30}$

Answer 1

The correct option is C $3\sqrt{30}$

Given lines $\overrightarrow{r}=\overrightarrow{a}+\alpha \left(\overrightarrow{b}\right)$
$\overrightarrow{r}=\overrightarrow{c}+\beta \left(\overrightarrow{d}\right)$
So, shortest distance $=\frac{\left|\right(\overrightarrow{a}-\overrightarrow{c}).(\overrightarrow{b}\times \overrightarrow{d}\left)\right|}{|\overrightarrow{b}\times \overrightarrow{d}|}$
Now, $\overrightarrow{b}\times \overrightarrow{d}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& -1& 1\\ -3& 2& 4\end{array}\right|=-6\hat{i}-15\hat{j}+3\hat{k}$

$\overrightarrow{a}-\overrightarrow{c}=6\hat{i}+15\hat{j}-3\hat{k}$

Shortest Distance$=\frac{|-36-225-9|}{3\sqrt{30}}$
$=3\sqrt{30}$