Question

The distance of the point A(–2, 3, 1) from the line PQ through P(–3, 5, 2) which make 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

The correct option is B $\sqrt{\frac{14}{3}}$

$Here,\alpha =\beta =\gamma \because cos2\alpha +cos2\beta +cos2\gamma =1\therefore cos\alpha =\frac{1}{\sqrt{3}}D{C}^{\prime }sofPQare(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$

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