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
$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 1 \\ 3 & 1 & 2 \end{vmatrix}=\hat{i}-\hat{j}-\hat{k}$
For the point on the line, putting $z=0,$ we get $\left.\begin{array}{l}2 x+y=1 \\ 3 x+y=2\end{array}\right\} x=1, y=-1$
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, $\overset{ \rightarrow }{a}=\hat{i}-\hat{j}$
$\overset{ \rightarrow }{b}=\hat{i}-\hat{j}-\hat{k}$
$\overset{ \rightarrow }{c}=\overset{ \rightarrow }{0}$
$\overset{ \rightarrow }{d}=\hat{i}-\hat{j}-\hat{k}$
$\overset{ \rightarrow }{c}-\overset{ \rightarrow }{a}=-\hat{i}+\hat{j}$
$\overset{ \rightarrow }{b}\times \overset{ \rightarrow }{d}=-2\hat{j}+2\hat{k}$
Shortest distance $=\left|\frac{\left(\overset{ \rightarrow }{c} - \overset{ \rightarrow }{a}\right) \cdot \left(\overset{ \rightarrow }{b} \times \overset{ \rightarrow }{d}\right)}{\left|\overset{ \rightarrow }{b} \times \overset{ \rightarrow }{d}\right|}\right|$
$=\left|\frac{- 2}{2 \sqrt{2}}\right|=\frac{1}{\sqrt{2}}$