question_answer

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

A) \[2/\sqrt{3}\]

B) \[\sqrt{14/3}\]

C) \[16/\sqrt{3}\]

D) \[5/\sqrt{3}\]

Correct Answer: B

Solution :

[b] Here, \[\theta =\beta =\gamma \] \[\therefore {{\cos }^{2}}\alpha +cs{{o}^{2}}\beta +{{\cos }^{2}}\gamma =1\]\[\therefore \cos \alpha =\frac{1}{\sqrt{3}}\] Direction cosines of PQ are \[\left( \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}} \right)\] \[PM=\]Projection of AP on PQ \[=\left| (-2+3)\frac{1}{\sqrt{3}}+(3-5)\frac{1}{\sqrt{3}}+(1-2)\frac{1}{\sqrt{3}} \right|=\frac{2}{\sqrt{3}}\] And \[AP=\sqrt{{{(-2+3)}^{2}}+{{(3-5)}^{2}}+{{(1-2)}^{2}}}=\sqrt{6}\] \[AM=\sqrt{{{(AP)}^{2}}-{{(PM)}^{2}}}=\sqrt{6-\frac{4}{3}}=\sqrt{\frac{14}{3}}\]