Question

The distance of the 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

Answer: (B)

Here, $\alpha =\beta =\gamma$
$co{s}^2\alpha +co{s}^2\beta +co{s}^2\gamma =1$
$\therefore cos\alpha =\frac{1}{\sqrt{3}}$
$D{C}^{\prime }s$ of $PQ$ are
$\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)$

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