First slide

Straight lines in 3D

Question

The distance of a point $A (- 2, 3, 1)$ from the line $\vec{M N}$ through $M (- 3, 5, 2)$ , which makes equal angles with the coordinate axes is

Moderate

A

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

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

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

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

Solution

The equation of line passing through $M (- 3, 5, 2)$ and making equal angles with coordinate axes is $\frac{x + 3}{1} = \frac{y - 5}{1} = \frac{z - 2}{1}$
General point on the above line is $P (k - 3, k + 5, k + 2)$
If the perpendicular distance from   $A (- 2, 3, 1)$ to the above line is $A P$   then   $A P$ is perpendicular to the line   $\frac{x + 3}{1} = \frac{y - 5}{1} = \frac{z - 2}{1}$
Hence,
$\begin{matrix} (k - 1) \cdot 1 + (k + 2) \cdot 1 + (k + 1) \cdot 1 = 0 \\ 3 k + 2 = 0 \\ k = - \frac{2}{3} \end{matrix}$
Hence the point   $P (k - 3, k + 5, k + 2)$ is $P (- \frac{11}{3}, - \frac{13}{5}, - \frac{4}{3})$
The required distance is $A P$ ,   $A P = \sqrt{\frac{25}{9} + \frac{16}{9} + \frac{1}{9}} = \sqrt{\frac{14}{3}}$

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