Question

The distance between the line
$\overline{r}=2\overline{i}-2\overline{j}+3\overline{k}+\lambda (\overline{i}-\overline{j}+4\overline{k})$ and the plane $\overline{r}.(\overline{i}+5\overline{j}+\overline{k})=5$ is

• A. $\frac{3}{10}$
• B. $\frac{10}{3}$
• C. $\frac{10}{3\sqrt{3}}$
• D. $\frac{10}{2\sqrt{3}}$

Answer 1

The correct option is C $\frac{10}{3\sqrt{3}}$

Given line is parallel to $\hat{i}-\hat{j}+4\hat{k}$

and plane has normal,$\overrightarrow{n}=\hat{i}+5\hat{j}+\hat{k}$

Taking dot product, $(\hat{i}+\hat{j}+4\hat{k}).(\hat{i}+5\hat{j}+\hat{k})=0$

$\Rightarrow$ line is parallel to plane ($\perp$ or to normal)

$\Rightarrow$ Distance between any point on line to

plane is same as that between line and

plane.

(2,-2,3) is a point on line.

equation of plane : x + 5y + z = 5

$\Rightarrow Distance=\left|\frac{2-10+3-5}{\sqrt{1+25+1}}\right|=\frac{10}{\sqrt{27}}=\frac{10}{3\sqrt{3}}$

$\Rightarrow$ No option.