Question

The shortest distance betwwen lines $\overline{r}=2\overline{i}-\overline{j}+\lambda (2\overline{i}+\overline{j}-3\overline{k})$ and $\overline{r}=\overline{i}-\overline{j}+2\overline{k}+\mu (2\overline{i}+\overline{j}-5\overline{k})$

• A. $6$
• B. $\frac{\sqrt{6}}{6}$
• C. $\frac{\sqrt{5}}{5}$
• D. $1$

Answer 1

The correct option is C $\frac{\sqrt{5}}{5}$

the given line (1) has error the eq should be

$\overrightarrow{r}=2\hat{i}-\hat{j}+\lambda (2\hat{i}+\hat{j}-5\hat{k})$

in place of

$\overrightarrow{r}=2\hat{i}-\hat{j}+\lambda (2\hat{i}+\hat{j}-3\hat{k})$

Given lines

$\overrightarrow{r}=2\hat{i}-\hat{j}+\lambda (2\hat{i}+\hat{j}-5\hat{k})$------(1)

$\overrightarrow{r}=\hat{i}-\hat{j}+2\hat{k}+\mu (2\hat{i}+\hat{j}-5\hat{k})$------(2)

the line sare parallel with normal vector

$\overrightarrow{b}=2\hat{i}+\hat{j}-5\hat{k}$

position vectors of line(1) and (2) respectively

$\overrightarrow{a}=2\hat{i}-\hat{j}$

and

$\overrightarrow{c}=\hat{i}-\hat{j}+2\hat{k}$

$\overrightarrow{c}-\overrightarrow{a}=\hat{i}-\hat{j}+2\hat{k}-(2\hat{i}-\hat{j})$

$\overrightarrow{c}-\overrightarrow{a}=\hat{i}-\hat{j}+2\hat{k}-2\hat{i}+\hat{j}$

$\overrightarrow{c}-\overrightarrow{a}=-\hat{i}+2\hat{k}$

shortest distance

$SD=\frac{\left|\right(\overrightarrow{c}-\overrightarrow{a})\times \overrightarrow{b}|}{\left|\overrightarrow{b}\right|}$

$SD=\frac{\left|\right(-\hat{i}+2\hat{k})\times (2\hat{i}+\hat{j}-5\hat{k}\left)\right|}{\sqrt{2^2+1^2+(-5{)}^2}}$

$SD=\frac{\left|\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -1& 0& 2\\ 2& 1& -5\end{array}\right|\right|}{\sqrt{2^2+1^2+(-5{)}^2}}$

$SD=\frac{\left|\hat{i}\right(0-2)-\hat{j}(5-4)+\hat{k}(-1-0\left)\right|}{\sqrt{4+1+25}}$

$SD=\frac{|-2\hat{i}-\hat{j}-\hat{k}|}{\sqrt{30}}$

$SD=\frac{\sqrt{(-2{)}^2+(-1{)}^2+(-1{)}^2}}{\sqrt{30}}$

$SD=\frac{\sqrt{6}}{\sqrt{30}}$

$SD=\frac{\sqrt{1}}{\sqrt{5}}$

on multiplying and dividing by $5$ in RHS

$SD=\frac{1\times 5}{\sqrt{5}\times 5}$

$SD=\frac{\sqrt{5}}{5}$