question_answer

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

A) \[\frac{10}{3\sqrt{3}}\]                  

B) \[\frac{10}{3}\]

C) \[\frac{10}{9}\]                                  

D) None of these

Correct Answer: A

Solution :

Given line is, \[\vec{r}=(2\hat{i}-2\hat{j}+3\hat{k})+\lambda (\hat{i}-\hat{j}+4\hat{k})\]         .....(1)

and given plane  \[\vec{r}.(\hat{i}+5\hat{j}+\hat{k})=5\]          ?..(2)

\[By\,(2),\,\,\vec{n}=(\hat{i}+5\hat{j}+\hat{k})\]

\[\because \,\,\vec{b}.\vec{n}=0\]

Therefore, the line is parallel to the plane. Thus, the distance between the line and the plane is equal to the length of the perpendicular from a point \[\vec{a}=(2\hat{i}-2\hat{j}+3\hat{k})\]on the line to the given plane.

Hence, the required distance \[=\left| \frac{(2\hat{i}-2\hat{j}+3\hat{k}).(\hat{i}+5\hat{j}+\hat{k})-5}{\sqrt{1+{{5}^{2}}+1}} \right|=\frac{10}{3\sqrt{3}}\]