Question

The shortest distance between lines$\frac{x-1}{2}=\frac{y+1}{0}=\frac{z}{1}$ and $\frac{x-2}{2}=\frac{y+1}{0}=\frac{z}{1}$

• A. $\frac{1}{\sqrt{5}}$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $\frac{1}{\sqrt{5}}$

Given eq of lines

$\frac{x-1}{2}=\frac{y+1}{0}=\frac{z}{1}$-----(1)

$\frac{x-2}{2}=\frac{y+1}{0}=\frac{z}{1}$-----(2)

position vector of line (1)

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

position vector of line (2)

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

normal vector

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

shortest between two parallel line

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

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

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

$AC=\hat{i}$

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

$\left|\overrightarrow{n}\right|=\sqrt{5}$

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

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

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

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

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

$SD=\frac{|-\hat{j}|}{\sqrt{5}}$

$SD=\frac{\sqrt{(-1{)}^2}}{\sqrt{5}}$

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