Question

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

• A. $\frac{10}{3\sqrt{3}}$
• B. $\frac{10}{9}$
• C. $\frac{10}{3}$
• D. $\frac{3}{10}$

Answer 1

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

The vector along normal to the plane is $\hat{i}+5\hat{j}+\hat{k}=\overrightarrow{n}$ say
The vector along the line is $\overrightarrow{b}=\hat{i}-\hat{j}+4\hat{k}$ and $\overrightarrow{n}.\overrightarrow{b}=(\hat{i}+5\hat{j}+\hat{k}).(\hat{i}-\hat{j}+4\hat{k})$
$=1-5+4=0$
$\therefore$ The line is parallel to the plane .The distance of the line from the plane is the distance of the point $(2,-2,3)$ from the plane $\overrightarrow{r}.(\hat{i}+5\hat{j}+\hat{k})=5$
i.e,$x+5y+z=5$
$\Rightarrow \frac{|2-10+3-5|}{\sqrt{1^2+5^2+1}}=\frac{10}{\sqrt{27}}=\frac{10}{3\sqrt{3}}$