First slide

Straight lines in 3D

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

Moderate

A

$\frac{2}{\sqrt{3}}$
B

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

$\frac{16}{\sqrt{3}}$
D

$\frac{5}{\sqrt{3}}$

Solution

Here, $α = β = γ$
$\begin{array}{ll} ∵ \cos^{2} α + \cos^{2} β + \cos^{2} γ = 1 \\ ∴ \cos α = \frac{1}{\sqrt{3}} \end{array}$
Direction cosines of PQ a $(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}})$
PM - Projection of AP on PQ
$\begin{array}{l} = |(- 2 + 3) \frac{1}{\sqrt{3}} + (3 - 5) \frac{1}{\sqrt{3}} + (1 - 2) \frac{1}{\sqrt{3}}| \\ = \frac{2}{\sqrt{3}} \\ and AP = \sqrt{(- 2 + 3)^{2} + (3 - 5)^{2} + (1 - 2)^{2}} = \sqrt{6} \\ AM = \sqrt{(AP)^{2} - (PM)^{2}} = \sqrt{6 - \frac{4}{3}} = \sqrt{\frac{14}{3}} \end{array}$

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