Question

The shortest distance between the skew lines
$\overrightarrow{r}=(\hat{i}+2\hat{j}+3\hat{k})+t(\hat{i}+3\hat{j}+2\hat{k})$ and $\overrightarrow{r}=(4\hat{i}+5\hat{j}+6\hat{k})+t(2\hat{i}+3\hat{j}+\hat{k})$ is

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

Answer 1

The correct option is D $\sqrt{3}$

Shortest distance $=\left|\frac{(\overrightarrow{a_2}-\overrightarrow{a_1}).({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2)}{|\overrightarrow{b_1}\times \overrightarrow{b_2}|}\right|$

$(\overrightarrow{b_1}\times \overrightarrow{b_2})=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 3& 2\\ 2& 3& 1\end{array}\right|=-3\hat{i}+3\hat{j}-3\hat{k}$

$|\overrightarrow{b_1}\times \overrightarrow{b_2}|=\sqrt{9+9+9}=3\sqrt{3}$

$\therefore$ shortest distance $=\left|\frac{(3\hat{i}+3\hat{j}+3\hat{k})\cdot (-3\hat{i}+3\hat{j}-3\hat{k})}{3\sqrt{3}}\right|$

$=\frac{3^2}{3\sqrt{3}}|(-1+1-1)|$

$=\sqrt{3}$