Question

The shortest distance between the lines
$\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}\mathrm{and},\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}$is
(a) $\sqrt{30}$
(b) $2\sqrt{30}$
(c) $5\sqrt{30}$
(d) $3\sqrt{30}$

Answer 1

(d) $3\sqrt{30}$

We have

$$\begin{gathered} \frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}...\left(1\right) \\ \frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}...\left(2\right) \end{gathered}$$

We know that line (1) passes through the point (3, 8, 3) and has direction ratios proportional to 3, $-$1, 1.

Its vector equation is $\overrightarrow{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1},\mathrm{where}\overrightarrow{a_1}=3\hat{i}+8\hat{j}+3\hat{k}\mathrm{and}\overrightarrow{b_1}=3\hat{i}-\hat{j}+\hat{k}.$

Also, line (2) passes through the point ($-$3, $-$7, 6) and has direction ratios proportional to $-$3, 2, 4.

Its vector equation is $\overrightarrow{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2},\mathrm{where}\overrightarrow{a_2}=-3\hat{i}-7\hat{j}+6\hat{k}\mathrm{and}\overrightarrow{b_2}=-3\hat{i}+2\hat{j}+4\hat{k}.$

Now,

$$\begin{gathered} \overrightarrow{a_2}-\overrightarrow{a_1}=-6\hat{i}-15\hat{j}+3\hat{k} \\ \overrightarrow{b_1}\times \overrightarrow{b_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& -1& 1\\ -3& 2& 4\end{array}\right| \\ =-6\hat{i}-15\hat{j}+3\hat{k} \\ \Rightarrow \left|\overrightarrow{b_1}\times \overrightarrow{b_2}\right|=\sqrt{{\left(-6\right)}^2+{\left(-15\right)}^2+3^2} \\ =\sqrt{36+225+9} \\ =\sqrt{270} \\ \left(\overrightarrow{a_2}-\overrightarrow{a_1}\right).\left(\overrightarrow{b_1}\times \overrightarrow{b_2}\right)=\left(-6\hat{i}-15\hat{j}+3\hat{k}\right).\left(-6\hat{i}-15\hat{j}+3\hat{k}\right) \\ =36+225+9 \\ =270 \end{gathered}$$

The shortest distance between the lines $\overrightarrow{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1}\mathrm{and}\overrightarrow{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}$ is given by

$$\begin{gathered} d=\left|\frac{\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right).\left(\overrightarrow{b_1}\times \overrightarrow{b_2}\right)}{\left|\overrightarrow{b_1}\times \overrightarrow{b_2}\right|}\right| \\ =\left|\frac{270}{\sqrt{270}}\right| \\ =\sqrt{270} \\ =3\sqrt{30} \end{gathered}$$