Vector Algebra Question 93
Question: Angle between the line $ \mathbf{r}=(\mathbf{i}+2\mathbf{j}-\mathbf{k})+\lambda (\mathbf{i}-\mathbf{j}+\mathbf{k}) $ and the normal to the plane $ \mathbf{r},.,(2\mathbf{i}-\mathbf{j}+\mathbf{k})=4 $ is
[MP PET 1997]
Options:
A) $ {{\sin }^{-1}},( \frac{2\sqrt{2}}{3} ) $
B) $ {{\cos }^{-1}},( \frac{2\sqrt{2}}{3} ) $
C) $ {{\tan }^{-1}},( \frac{2\sqrt{2}}{3} ) $
D) $ {{\cot }^{-1}},( \frac{2\sqrt{2}}{3} ) $
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Answer:
Correct Answer: A
Solution:
- The plane is $ 2x-y+z=4 $ and the line is $ \frac{x-1}{1}=\frac{y-2}{-1}=\frac{z+1}{1} $
$ \therefore ,\sin \theta =\frac{2+1+1}{\sqrt{6}\sqrt{3}}=\frac{4}{\sqrt{18}}=\frac{2\sqrt{2}}{3}\Rightarrow \theta ={{\sin }^{-1}}( \frac{2\sqrt{2}}{3} ) $ .
BETA
