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

• 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, $\alpha =\beta =\gamma$
$\because {\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma =1$
$\therefore \cos \alpha =\frac{1}{\sqrt{3}}$
Direction cosines of $PQ$ are $(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$

$PM=$ Projection of $AP$ on $PQ$
$=\left|\right(-2+3)\frac{1}{\sqrt{3}}+(3-5)\frac{1}{\sqrt{3}}+(1-2\left)\frac{1}{\sqrt{3}}\right|$
$=\frac{2}{\sqrt{3}}$
and $AP=\sqrt{(-2+3{)}^2+(3-5{)}^2+(1-2{)}^2}=\sqrt{6}$
$AM=\sqrt{(AP{)}^2-(PM{)}^2}=\sqrt{6-\frac{4}{3}}=\sqrt{\frac{14}{3}}$