Vector Algebra Question 114
Question: The distance of the point $ 2\mathbf{i}+\mathbf{j}-\mathbf{k} $ from the plane $ \mathbf{r}.(\mathbf{i}-2\mathbf{j}+4\mathbf{k})=9 $ is
Options:
A) $ \frac{13}{\sqrt{21}} $
B) $ \frac{13}{21} $
C) $ \frac{13}{3\sqrt{21}} $
Answer:
Correct Answer: A
Solution:
- We know that the perpendicular distance of a point P with position vector $ \mathbf{a} $ from the plane $ \mathbf{r}.,\mathbf{n}=d $ is given by $ \frac{|\mathbf{a}.\mathbf{n}-d|}{|\mathbf{n}|} $ . Here $ \mathbf{a}=2\mathbf{i}+\mathbf{j}-\mathbf{k},\mathbf{n}=\mathbf{i}-2\mathbf{j}+4\mathbf{k} $ and $ d=9 $ . So, required distance $ =\frac{|(2\mathbf{i}+\mathbf{j}-\mathbf{k}).(\mathbf{i}-2\mathbf{j}+4\mathbf{k})-9|}{\sqrt{1+4+16}} $ $ =\frac{|2-2-4-9|}{\sqrt{21}}=\frac{13}{\sqrt{21}} $ . $ \frac{3}{\sqrt{21}} $
BETA
