Three Dimensional Geometry Question 359
Question: The distance of point A (-2, 3, 1) from the line PQ through P (- 3, 5, 2), which makes equal angles with the axes is
Options:
A) $ 2/\sqrt{3} $
B) $ \sqrt{14/3} $
C) $ 16/\sqrt{3} $
D) $ 5/\sqrt{3} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Here, $ \theta =\beta =\gamma $
$ \therefore {{\cos }^{2}}\alpha +cso^{2}\beta +{{\cos }^{2}}\gamma =1 $
$ \therefore \cos \alpha =\frac{1}{\sqrt{3}} $ Direction cosines of PQ are $ ( \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}} ) $ $ PM= $ Projection of AP on PQ $ =| (-2+3)\frac{1}{\sqrt{3}}+(3-5)\frac{1}{\sqrt{3}}+(1-2)\frac{1}{\sqrt{3}} |=\frac{2}{\sqrt{3}} $ And $ AP=\sqrt{{{(-2+3)}^{2}}+{{(3-5)}^{2}}+{{(1-2)}^{2}}}=\sqrt{6} $ $ AM=\sqrt{{{(AP)}^{2}}-{{(PM)}^{2}}}=\sqrt{6-\frac{4}{3}}=\sqrt{\frac{14}{3}} $
BETA
