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. $2/\sqrt{3}$
• B. $\sqrt{14/3}$
• C. $16/\sqrt{3}$
• D. $5/\sqrt{3}$

Answer 1

Answer: (B)

$\sqrt{14/3}$

Given,

The line $PQ$ makes equal angle with axes, say $\alpha$.

So the direction cosine of the line will be $(\cos \alpha ,\cos \alpha ,\cos \alpha )$

Also,

$\cos {}^2\alpha +\cos {}^2\alpha +\cos {}^2\alpha =1\cos \alpha =\frac{1}{\sqrt{3}}$

$\therefore$ the direction cosine of the line is $(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$

Next,we move on to find the projection of $AP$ on $PQ$

$PD=(-2+3)\frac{1}{\sqrt{3}}+(3-5)\frac{1}{\sqrt{3}}+(1-2)\frac{1}{\sqrt{3}}=\frac{-2}{\sqrt{3}}\therefore {AD}^2={AP}^2-{PD}^2={(3-2)}^2+{(3-5)}^2+{(1-2)}^2-{\left(\frac{2}{\sqrt{3}}\right)}^2=6-\frac{4}{3}=\frac{14}{3}\therefore AD=\sqrt{\frac{14}{3}}$