A) \[\frac{10}{3}\]
B) \[\frac{10}{\sqrt{3}}\]
C) \[\frac{10}{3\sqrt{3}}\]
D) \[\frac{10}{9}\]
Correct Answer: C
Solution :
We have the given line \[\overrightarrow{r}=2\hat{i}-2\hat{j}+3\hat{k}+\lambda (\hat{i}-\hat{j}+4\hat{k})\] On comparing it with\[\overrightarrow{r}=\overrightarrow{a}+t\,\overrightarrow{b},\]we get \[\overrightarrow{a}=2\hat{i}-2\hat{j}+3\hat{k},\]\[\overrightarrow{b}=\hat{i}-\hat{j}+4\hat{k}\] Also, the plane is \[\overrightarrow{r}\cdot (\hat{i}+5\hat{j}+\hat{k})=5\] On comparing it with \[\overrightarrow{r}\cdot \overrightarrow{n}=d,\]we get \[\overrightarrow{n}=\hat{i}+5\hat{j}+\hat{k}\] and \[d=5\] Since, \[\overrightarrow{b}\cdot \overrightarrow{n}=(\hat{i}-\hat{j}+4\hat{k})\cdot (\hat{i}+5\hat{j}+\hat{k})\] \[=1-5+4=0\] \[\therefore \]Given line is parallel to the given plane. Now, distance between the line and the plane is given by required distance\[=\left| \frac{\overrightarrow{a}\cdot \overrightarrow{n}-d}{\left| \overrightarrow{n} \right|} \right|\] \[=\frac{\left| (2\hat{i}-2\hat{j}+3\hat{k})\cdot (\hat{i}+5\hat{j}+\hat{k})-5 \right|}{\sqrt{1+25+1}}\] \[=\frac{\left| 2-10+3-5 \right|}{\sqrt{1+25+1}}=\frac{10}{3\sqrt{3}}\]
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