A) 5
B) 4
C) 3
D) 2
E) 1
Correct Answer: D
Solution :
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 \[=\left| \frac{(2\hat{i}+2\hat{j}-\hat{k}).(\hat{i}+2\hat{j}+2\hat{k})-10}{\sqrt{1+4+4}} \right|\] \[=\left| \frac{2+4-2-10}{\sqrt{9}} \right|\] \[=\left| \frac{-6}{3} \right|=2\]
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