Question

Find the shortest distance between the lines $\overline{r}=4\overline{i}-\overline{j}+\lambda (\overline{i}+2\overline{j}-5\overline{k})$ and $\overline{r}=\overline{i}-\overline{j}+2\overline{k}+\mu (\overline{i}+2\overline{j}-5\overline{k})$

• A. $\sqrt{220}$
• B. $\frac{\sqrt{221}}{\sqrt{30}}$
• C. $432$
• D. $\sqrt{33}$

Answer 1

Answer: (B)

$\frac{\sqrt{221}}{\sqrt{30}}$

Given eq of lines

$\overrightarrow{r}=4\hat{i}-\hat{j}+\lambda (\hat{i}+2\hat{j}-5\hat{k})$----(1)

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

position vector of line (1)

$\overrightarrow{a}=4\hat{i}-\hat{j}$

position vector of line (2)

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

normal vector

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

shortest between two parallel line

$SD=\frac{|\overrightarrow{AC}\times (\overrightarrow{n}\left)\right|}{\left|\overrightarrow{n}\right|}$

$AC=\overrightarrow{c}-\overrightarrow{a}$

$AC=\hat{i}-\hat{j}+2\hat{k}-(4\hat{i}-\hat{j})$

$AC=\hat{i}-\hat{j}+2\hat{k}-4\hat{i}+\hat{j}$

$AC=-3\hat{i}+2\hat{k}$

$\left|\overrightarrow{n}\right|=\sqrt{(1{)}^2+2^2+(-5{)}^2}$

$\left|\overrightarrow{n}\right|=\sqrt{30}$

putting AC $\overrightarrow{n},\left|\overrightarrow{n}\right|$ in formula

$SD=\frac{|\overrightarrow{AC}\times (\overrightarrow{n}\left)\right|}{\left|\overrightarrow{n}\right|}$

$SD=\frac{\left|\right(-3\hat{i}+2\hat{k})\times (\hat{i}+2\hat{j}-5\hat{k}\left)\right|}{\sqrt{30}}$

$SD=\frac{\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -3& 0& 2\\ 1& 2& -5\end{array}\right|}{\sqrt{30}}$

$SD=\frac{\left|\hat{i}\right(0-4)-\hat{j}(15-2)+\hat{k}(-6-0\left)\right|}{\sqrt{30}}$

$SD=\frac{|-4\hat{i}-13\hat{j}-6\hat{k}|}{\sqrt{30}}$

$SD=\frac{\sqrt{(-4{)}^2+(-13{)}^2+(-6{)}^2}}{\sqrt{30}}$

$SD=\frac{\sqrt{16+169+36}}{\sqrt{30}}$

$SD=\frac{\sqrt{221}}{\sqrt{30}}$