Find an equation of the plane that passes through the points $(0-2,5)$ and $(-1,3,1)$ and is perpendicular to the plane $2z = 5x + 4y$.
Here's what I have so far:
The plane through $(0,-2,5)$ is $ax + b(y+z) + c(z-5) = 0$.
And the plane also passes through $(-1,3,1)$ so I get: $$-a + 5b – 4c = 0 \tag{1}.$$
When I looked at the explanation it says:
Now we know that the plane is perpendicular to $5x + 4y – 2z = 0$ and then it replaces $(x,y,z)$ with $(a,b,c)$ to get $$5a + 4b – 2c = 0. \tag{2}$$
It continues from there saying to solve the two equations to get $\frac{a}{6} = \frac{b}{-22} = \frac{c}{-29}$.
I know how to solve it once it gets to this but I have absolutely no idea how they got to this step.
Best Answer
First take a direction vector to those two points, then, take the given normal from the equation of a plane. The cross product of the direction vector with the normal will then gives the normal of the desired plane, finally take either of the two point and form a point normal equation of a plane.