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