Question

The shortest distance between the line $x=y=z$ and the line of intersection of $2x+y+z-1=0$ and $3x+y+2z-2=0$ is

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

Answer 1

Answer: (A)

$P_1=2x+y+z-1=0$
$P_2=3x+y+2z-2=0$
The line of intersection is parallel to
$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& 1\\ 3& 1& 2\end{array}\right|=\hat{i}-\hat{j}-\hat{k}$
For the point on the line, putting $z=0,$ we get $\begin{array}{l}2x+y=1\\ 3x+y=2\end{array}\}x=1,y=-1$
So, the required point is $\left(1,-1,0\right)$
Hence, the equation of the line of intersection is
$\frac{x-1}{1}=\frac{y+1}{-1}=\frac{z-0}{-1}$
Let, $\overrightarrow{a}=\hat{i}-\hat{j}$
$\overrightarrow{b}=\hat{i}-\hat{j}-\hat{k}$
$\overrightarrow{c}=\overrightarrow{0}$
$\overrightarrow{d}=\hat{i}-\hat{j}-\hat{k}$
$\overrightarrow{c}-\overrightarrow{a}=-\hat{i}+\hat{j}$
$\overrightarrow{b}\times \overrightarrow{d}=-2\hat{j}+2\hat{k}$
Shortest distance $=\left|\frac{\left(\overrightarrow{c}-\overrightarrow{a}\right)\cdot \left(\overrightarrow{b}\times \overrightarrow{d}\right)}{\left|\overrightarrow{b}\times \overrightarrow{d}\right|}\right|$
$=\left|\frac{-2}{2\sqrt{2}}\right|=\frac{1}{\sqrt{2}}$