Question

The distance of the point A (- 2, 3, 1) from the line PQ through P (-3, 5, 2) which make equal angles with the axes is

• A. $\frac{2}{\sqrt{3}}$
• B. $\sqrt{\frac{14}{3}}$
• C. $\frac{16}{\sqrt{3}}$
• D. $\frac{5}{\sqrt{3}}$

Answer 1

The correct option is B $\sqrt{\frac{14}{3}}$

Here, $\alpha =\beta =\gamma$
$\because co{s}^2\alpha +co{s}^2\beta +co{s}^2\gamma =1$
$\therefore cos\alpha =\frac{1}{\sqrt{3}}$
DC's 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 $AP=\sqrt{(-2+3{)}^2+(3-5{)}^2}=\sqrt{6}$
$AM=\sqrt{(AP{)}^2-(PM{)}^2}=\sqrt{6-\frac{4}{3}}=\sqrt{\frac{14}{3}}$