Question

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

• A. $\frac{10}{3\sqrt{3}}$
• B. $\frac{10}{3}$
• C. $\frac{10}{9}$
• D. $none$

Answer 1

The correct option is D $none$

$\begin{array}{c}\overrightarrow{r}=2\hat{i}-2\hat{j}+3\hat{k}+\lambda \left(\hat{i}-\hat{j}+4\hat{k}\right)\\ \frac{x-2}{1}=\frac{y+2}{-1}=\frac{z-3}{4}\\ \overrightarrow{r}.\left(\hat{i}+5\hat{j}+\hat{k}\right)=5\\ x+5y+z=5\\ \sin \theta =\frac{1-5+4}{\sqrt{18}\sqrt{27}}=0\end{array}$

Hence, line and plane are parallel.

Distance b/w line and plane.

Again,

$\frac{x-2}{1}=\frac{y+2}{1}=\frac{z-3}{4}$

$x+5y+z=5$

$d=\frac{\left|-2+5\left(2\right)-3\right|}{\sqrt{27}}=\frac{5}{\sqrt{27}}$

Hence length of perpendicular $=\frac{5}{\sqrt{27}}$

Which is distance b/w line and plane.

Then,

Option $D$ is correct answer.