Vector Algebra Question 353
Question: The distance from the point $ -\mathbf{i}+2\mathbf{j}+6\mathbf{k} $ to the straight line through the point (2, 3, ?4) and parallel to the vector $ 6\mathbf{i}+3\mathbf{j}-4\mathbf{k} $ is
Options:
A) 7
B) 10.
C) 9
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
We have $ \overrightarrow{AP}=-3\mathbf{i}-\mathbf{j}+10\mathbf{k} $ \ $ |\overrightarrow{AP}|=\sqrt{9+1+100}=\sqrt{110} $ $ AN= $ Projection of $ \overrightarrow{AP} $ on $ 6\mathbf{i}+3\mathbf{j}-4\mathbf{k} $ $ =| \frac{\overrightarrow{AP}.(6\mathbf{i}+3\mathbf{j}-4\mathbf{k})}{|6\mathbf{i}+3\mathbf{j}-4\mathbf{k}|} | $ $ =| \frac{-18-3-40}{\sqrt{61}} |=\frac{59}{\sqrt{61}} $ \ $ PN=\sqrt{AP^{2}-AN^{2}} $ $ =\sqrt{110-\left(\frac{59}{\sqrt{61}}\right)^2}=7 $ .
BETA
