Linear-Algebra - How to find a position vector of a point S on a plane, that is also perpendicular to the plane

Question

I want to find the coordinates of a position vector of a point S on a plane. This position vector must be perpendicular to the plane. I don't really have an idea on how to approach this.

The general form of the equation of the plane W is: x - 2y + z = -2. The answer must be in this form:

(s_x, s_y, s_z) with s_x, s_y and s_z as the x, y and z coordinates of the vector s

Thanks in advance

Answer 1

The plane

$$x-2y+z=-2$$

has normal vector $$\vec n=(1,-2,1)$$

thus the position vector we are looking for is in the form

$$(s_x,s_y,s_z)=a(1,-2,1)$$

by the condition to belong to the plane we obtain

$$a-2(-2a)+a=-2\iff6a=-2\iff a=-\frac13$$

therefore

$$(s_x,s_y,s_z)=\left(-\frac13,\frac23,-\frac13\right)$$