Question

The distance between the line $\overrightarrow{r}=(2\hat{i}+2\hat{j}-\hat{k})+\lambda (2\hat{i}+\hat{j}-2\hat{k})$ and the plane $\overrightarrow{r}.(\hat{i}+2\hat{j}+2\hat{k})=10$ is equal to

• A. 5
• B. 4
• C. 3
• D. 2

Answer 1

Answer: (D)

2

Answer 2

The given line is $\overrightarrow{r}=(2\hat{i}+2\hat{j}-\hat{k})+\lambda (2\hat{i}+\hat{j}+2\hat{k})$ or $\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$ where $\overrightarrow{a}=(2\hat{i}+2\hat{j}-\hat{k}),$ and $\overrightarrow{b}=2\hat{i}+\hat{j}-2\hat{k}$
The equation of plane is $\overrightarrow{r}.(\hat{i}+2\hat{j}+2\hat{k})=10$ or $\overrightarrow{r}.\hat{n}=d$ where $\hat{n}=(\hat{i}+2\hat{j}+2\hat{k})$ Since, $\hat{b}\,.\,\hat{n}=(2\hat{i}+\hat{j}-2\hat{k}).(\hat{i}+2\hat{j}+2\hat{k})$
$=2+2-4=0$
Therefore, the line is parallel to the plane. Hence, the required distance
$=\left| \frac{(2\hat{i}+2\hat{j}-\hat{k}).(\hat{i}+2\hat{j}+2\hat{k})-10}{\sqrt{1+4+4}} \right|$
$=\left| \frac{2+4-2-10}{\sqrt{9}} \right|$
$=\left| \frac{-6}{3} \right|=2$