Question

The straight line
$\overrightarrow{r}=(\hat{i}+\hat{j}+2\hat{k})+t(2\hat{i}+5\hat{j}+3\hat{k})$ is parallel to the plane $\overrightarrow{r}.(2\hat{i}+\hat{j}-3\hat{k})=5$. Then, the distance between the straight line and the plane is

• A. $\frac{9}{\sqrt{14}}$
• B. $\frac{8}{\sqrt{14}}$
• C. $\frac{7}{\sqrt{14}}$
• D. $\frac{6}{\sqrt{14}}$
• E. $\frac{5}{\sqrt{14}}$

Answer 1

Answer: (B)

$\frac{8}{\sqrt{14}}$

Line $\overrightarrow{r}=(\hat{i}+\hat{j}+2\hat{k})+t(2\hat{i}+5\hat{j}+3\hat{k})$
Plane $\overrightarrow{r}.(2\hat{i}+\hat{j}-3\hat{k})=5$
Since $\overrightarrow{n}=2\hat{i}+\hat{j}-3\hat{k}$ is normal to plane
$⟹|\overrightarrow{n}{|}^2=14$
So for some k $(\hat{i}+\hat{j}+2\hat{k})+k\ast \overrightarrow{n}$ will be on the plane
Distance between the plane is $|k\overrightarrow{n}|$
Solve for k by applying on plane above point
$\left(\right(\hat{i}+\hat{j}+2\hat{k})+k\overrightarrow{n}).\overrightarrow{n}=5$
$(\hat{i}+\hat{j}+2\hat{k}).\overrightarrow{n}+k|\overrightarrow{n}{|}^2=5$
$2+1-6+14k=5⟹k=\frac{8}{14}$
Distance between the Line and plane is $|k\overrightarrow{n}|=\frac{8}{14}\sqrt{14}=\frac{8}{\sqrt{14}}$