Question

The distance of the point $A(-2,3,1)$ from the line $PQ$ through $P(-3,5,2)$ which make equal angles with 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$\alpha =\beta =\gamma \Rightarrow {\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma =1$
$\cos \alpha =\frac{1}{\sqrt{3}}$
$A(-2,3,1)\text{and}P(-3,5,2)\Rightarrow AP=\sqrt{6}$
d.c's of line $PQ:(l,m,n)=(-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}})$ or $(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$
$PM=\text{projection of}\overrightarrow{AP}\text{on}\overrightarrow{PQ}=|\overrightarrow{AP}\cdot (l\hat{i}+m\hat{j}+n\hat{k})|=\frac{2}{\sqrt{3}}$
$\Rightarrow AM=\sqrt{(AP{)}^2-(PM{)}^2}=\sqrt{\frac{14}{3}}$