Three Dimensional Geometry Question 379
Question: 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
Options:
A) $ \frac{10}{3\sqrt{3}} $
B) $ \frac{10}{9} $
C) $ \frac{10}{3} $
D) $ \frac{3}{10} $
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Answer:
Correct Answer: A
Solution:
[a] It is obvious that the given line and plane are parallel. Given point on the lie is $ A(2,-2,3). $ $ B(0,0,5) $ is a point the plane. Therefore, $ \xrightarrow[AB]{}=(2-0)\hat{i}+(-2-0)\hat{j}+(3-5)\hat{k} $ Then distance of B from the plane = Projection of $ \xrightarrow[AB]{} $ on vector $ \hat{i}+5\hat{j}+\hat{k} $ $ P=| \frac{(2\hat{i}-2\hat{j}-2\hat{k}).(\hat{i}+5\hat{j}+\hat{k})}{\sqrt{1+25+1}} |=| \frac{2-10-2}{\sqrt{27}} |=\frac{10}{3\sqrt{3}} $
BETA
