Question

The distance of point $A(-2,3,1)$ from the 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)

$\sqrt{\frac{14}{3}}$

Since $\alpha =\beta =\gamma$
$\Rightarrow l=m=n=\frac{1}{\sqrt{3}}$, where $l,m,n$ are direction cosines of line $PQ$
Let $M$ be a point on the line $PQ$ such that $AM\perp PQ$
So, $PM$ $=$ Projection of $AP$ on $PQ$

$=\mid (-2+3)\frac{1}{\sqrt{3}}+(3-5)\frac{1}{\sqrt{3}}+(1-2)\frac{1}{\sqrt{3}}\mid =\frac{2}{\sqrt{3}}$

and $AP=\sqrt{(-2+3{)}^2+(3-5{)}^2+(1-2{)}^2}=\sqrt{6}$
Hence, required distance is,

$AM=\sqrt{P{Q}^2-Q{M}^2}=\sqrt{\frac{14}{3}}$