Question

The distance of the point $(1,3,–7)$ from the plane passing through the point $(1,–1,–1),$ having normal perpendicular to both the lines $\frac{x-1}{1}=\frac{y+2}{-2}=\frac{z-4}{3}$ and $\frac{x-2}{2}=\frac{y+1}{-1}=\frac{z+7}{-1}$, is:

• A. $\frac{20}{\sqrt{74}}$
• B. $\frac{10}{\sqrt{83}}$
• C. $\frac{5}{\sqrt{83}}$
• D. $\frac{10}{\sqrt{74}}$

Answer 1

The correct option is B $\frac{10}{\sqrt{83}}$

As the plane is parallel to both the lines. So the cross product of direction ratios of the lines will give the normal vector of the plane
$(1,-2,3)\times (2,-1,-1)=(5,7,3)$
The equation of plane with normal and a point is
$\overrightarrow{r}.\overrightarrow{n}=\overrightarrow{a}.\overrightarrow{n}(\overrightarrow{r}-\overrightarrow{a}).\overrightarrow{n}=0$.
So the equation of plane passing through the point $(1,-1,-1)$ and direction ratios $(5,7,3)$ is

$5(x–1)+7(y+1)+3(z+1)=0$
Hence, $5x+7y+3z+5=0$ is the equation of the plane. So, the distance of the point $P(1,3,–7)$ from the plane is
$d=\left|\frac{5+21-21+5}{\sqrt{25+49+9}}\right|=\left|\frac{10}{\sqrt{83}}\right|\text{units}$