The equation of a line which is perpendicular to the plane 5x+2y−8z=0 and passing through -1,0,1 is

# The equation of a line which is perpendicular to the plane $5x+2y-8z=0$ and passing through $\left(-1,0,1\right)$ is

1. A

$\frac{x+1}{5}=\frac{y}{2}=\frac{z-1}{-8}$

2. B

$\frac{x-1}{5}=\frac{y-1}{0}=\frac{z-1}{-8}$

3. C

$\frac{x+1}{8}=\frac{z-1}{2}=\frac{y-0}{2}$

4. D

$\frac{x-5}{-1}=\frac{y-2}{0}=\frac{z+2}{1}$

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### Solution:

The equation of a line passing through the point $\left({x}_{1},{y}_{1},{z}_{1}\right)$ and perpendicular to the plane $ax+by+cz+d=0$ is $\frac{x-{x}_{1}}{a}=\frac{y-{y}_{1}}{b}=\frac{z-{z}_{1}}{c}$

Therefore, the equation of a line which is perpendicular to the plane $5x+2y-8z=0$ and passing through $\left(-1,0,1\right)$ is $\frac{x+1}{5}=\frac{y}{2}=\frac{z-1}{-8}$

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