Question

The shortest distance between the lines$\overline{r}=3\overline{i}+5\overline{j}+7\overline{k}+\lambda (\overline{i}+2\overline{j}+\overline{k})$$\overline{r}=-\overline{i}-\overline{j}+\overline{k}+\mu (7\overline{i}-6\overline{j}+\overline{kk})$ is ?

• A. $\frac{116}{\sqrt{513}}$
• B. $\frac{26}{5\sqrt{5}}$
• C. $\frac{46}{5\sqrt{5}}$
• D. $\frac{36}{5\sqrt{5}}$

Answer 1

Answer: (A)

$\frac{116}{\sqrt{513}}$

Consider the problem

Let the given lines be

$r=a_1+\lambda b_2$

And $r=a_2+\lambda b_2$

Now, Shortest distance between two lines

$d=\left|\frac{\left({\overrightarrow{a}}_2-{\overrightarrow{a}}_1\right).\left({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2\right)}{\left|{\overrightarrow{b}}_1\times {\overrightarrow{b}}_2\right|}\right|$

where, $a_1=3\overline{i}+5\overline{j}+7\overline{k},a_2=-\overline{i}-\overline{j}+\overline{k}$

$b_1=\overline{i}+2\overline{j}+\overline{k},b_2=7\overline{i}-6\overline{j}+\overline{k}$

Therefore,

$a_2-a_1=-4\overline{i}+4\overline{j}-6\overline{k}$

${\overrightarrow{b}}_1\times {\overrightarrow{b}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 1\\ 7& -6& 1\end{array}\right|$

$=8\hat{i}+7\hat{j}-20\hat{k}$

$\left|{\overrightarrow{b}}_1\times {\overrightarrow{b}}_2\right|=\sqrt{8^2+7^2+\left(-{20}^2\right)}=\sqrt{64+49+400}=\sqrt{513}$

So,

$d=\left|\frac{-4\overline{i}+4\overline{j}-6\overline{k}.\left(8\hat{i}+7\hat{j}-20\hat{k}\right)}{\sqrt{513}}\right|$

$d=\left|\frac{-32+28+120}{\sqrt{513}}\right|=\frac{116}{\sqrt{513}}$