Question

The shortest distance between lines $\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}$ and $\frac{x-3}{7}=\frac{y-5}{-6}=\frac{z-7}{1}$ approximately is

• A. $24$
• B. $45$
• C. $\sqrt{116}$
• D. $39$

Answer 1

The correct option is C $\sqrt{116}$

Given eq of lines

$\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}$-----(1)

$\frac{x-3}{7}=\frac{y-5}{-6}=\frac{z-7}{1}$-----(2)

position vector of line (1)

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

position vector of line (2)

$\overrightarrow{c}=3\hat{i}+5\hat{j}+7\hat{k}$

normal vector

$\overrightarrow{n}=7\hat{i}-6\hat{j}+\hat{k}$

shortest between two parallel line

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

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

$AC=3\hat{i}+5\hat{j}+7\hat{k}-(-\hat{i}-\hat{j}-\hat{k})$

$AC=4\hat{i}+6\hat{j}+8\hat{k}$

$\left|\overrightarrow{n}\right|=\sqrt{(7{)}^2+(-6{)}^2+1^2}$

$\left|\overrightarrow{n}\right|=\sqrt{86}$

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

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

$SD=\frac{\left|\right(4\hat{i}+6\hat{j}+8\hat{k})\times (7\hat{i}-6\hat{j}+\hat{k}\left)\right|}{\sqrt{86}}$

$SD=\frac{\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 4& 6& 8\\ 7& -6& 1\end{array}\right|}{\sqrt{86}}$

$SD=\frac{\left|\hat{i}\right(6+48)-\hat{j}(4-56)+\hat{k}(-24-42\left)\right|}{\sqrt{86}}$

$SD=\frac{|54\hat{i}+52\hat{j}-66\hat{k}|}{\sqrt{86}}$

$SD=\frac{\sqrt{(54{)}^2+(52{)}^2+(-66{)}^2}}{\sqrt{86}}$

$SD=\frac{\sqrt{2916+2704+4356}}{\sqrt{86}}$

$SD=\frac{\sqrt{9976}}{\sqrt{86}}$

$SD=\sqrt{116}$