Question

The distance between the plane whose equation is $\overline{r}\cdot \left(2\overline{i}+\overline{j}-3\overline{k}\right)=5$ and the line whose equation is $\overline{r}=i+\lambda \left(2\overline{i}+5\overline{j}+3\overline{k}\right)$ is

Answer 1

From the question we get

$\overrightarrow{r}.\left(2\hat{i}+\hat{j}-3\hat{z}\right)=5$

$\Rightarrow 2x+y-3z=5$

And,

$\overrightarrow{r}=\hat{i}+\lambda \left(2\hat{i}+5\hat{j}-3\hat{k}\right)=5$

$\Rightarrow \frac{x-1}{2}=\frac{y-0}{5}=\frac{z-0}{3}$

$\therefore \left(2\right)\left(2\right)+5\left(1\right)-3\left(3\right)=0$

Line is $\parallel$ to plane

$\therefore$ Distance between plane and line$=$length of perpendicular from $(1,0,0)$ to plane.

$d=\frac{\left|2\left(1\right)+\left(0\right)-3\left(0\right)-5\right|}{\sqrt{{\left(2\right)}^2+{\left(1\right)}^2+{\left(-3\right)}^2}}=\frac{\left|-3\right|}{\sqrt{14}}$

$d=\frac{3}{\sqrt{14}}$

Which is the required answer.