Question

The distance between the line $x=2+t,$ $y=1+t,$ $z=\frac{-1}{2}-\frac{t}{2}$ where $t\in R$ and the plane $\overrightarrow{r}\cdot (\hat{i}+2\hat{j}+6\hat{k})=10$ is

• A. $\frac{1}{6}$
• B. $\frac{1}{\sqrt{41}}$
• C. $\frac{1}{7}$
• D. $\frac{9}{\sqrt{41}}$

Answer 1

The correct option is D $\frac{9}{\sqrt{41}}$

We have, Line : $\frac{x-2}{1}=\frac{y-1}{1}=\frac{z+\frac{1}{2}}{-1/2}=t$ (let)
Plane $:x+2y+6z-10=0$
Vector along the line, $\overrightarrow{v}=\hat{i}+\hat{j}-\frac{1}{2}\hat{k}$
Vector normal to the plane, $\overrightarrow{n}=\hat{i}+2\hat{j}+6\hat{k}$
As $\overrightarrow{v}\cdot \overrightarrow{n}=1+2-\frac{1}{2}\times 6=0,$
hence line is parallel to the plane.

Now distance from the point $(2,1,-\frac{1}{2})$ to the plane $x+2y+6z-10=0$ is
$d=\frac{|2+2+6(-\frac{1}{2})-10|}{\sqrt{36+4+1}}\Rightarrow d=\frac{9}{\sqrt{41}}$