Question

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

• A. $\sqrt{30}$
• B. $2\sqrt{30}$
• C. $5\sqrt{30}$
• D. $3\sqrt{30}$

Answer 1

Answer: (D)

$3\sqrt{30}$

We have $\overrightarrow{a_1}=3\hat{i}+8\hat{j}+3\hat{k}$ $\overrightarrow{b_1}=3\hat{i}-\hat{j}+\hat{k}$
$\overrightarrow{a_2}=-3\hat{i}-7\hat{j}+6\hat{k}$ and $\overrightarrow{b_2}=-3\hat{i}+2\hat{j}+4\hat{k}$
Now, $b_1\times b_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& -1& 1\\ -3& 2& 4\end{array}\right|=-6\hat{i}-15\hat{j}+3\hat{k}$
$\mid b_1\times b_2\mid =\sqrt{6^2+{15}^2+3^2}=3\sqrt{30}$
$a_1-a_2=6\hat{i}+15\hat{j}-3\hat{k}$
Therefore shortest distance between both the lines is
$d=\mid \frac{(a_1-a_2)\times (b_1\times b_2)}{\mid b_1\times b_2\mid }\mid =\frac{6^2+{15}^2+3^2}{3\sqrt{30}}=3\sqrt{30}$