First slide

Straight lines in 3D

Question

Equation of the line passing through the point $(- 1, 2, - 3)$ and perpendicular to the plane $2 x + 3 y + z + 5 = 0$

Easy

A

$\frac{x - 1}{- 1} = \frac{y + 2}{1} = \frac{z - 3}{- 1}$
B

$\frac{x + 1}{- 1} = \frac{y - 2}{1} = \frac{z + 3}{1}$
C

$\frac{x + 1}{2} = \frac{y - 2}{3} = \frac{z + 3}{1}$
D

$\frac{x + y}{- 3} = \frac{y - 1}{2} = \frac{z - 3}{1}$

Solution

The direction ratios of any line perpendicular to the plane $a x + b y + c z + d = 0$ are proportional to $〈a, b, c〉$
The equation of the line passing through the point $(x_{1}, y_{1}, z_{1})$ and parallel to the line whose direction ratios are $〈a, b, c〉$ is $\frac{x - x_{1}}{a} = \frac{y - y_{1}}{b} = \frac{z - z_{1}}{c}$
Hence, the equation of the line passing through the point $(- 1, 2, - 3)$ and perpendicular to the plane $2 x + 3 y + z + 5 = 0$ is $\frac{x + 1}{2} = \frac{y - 2}{3} = \frac{z + 3}{1}$

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