Suppose that I want to find the equation of the plane that is perpendicular to the plane $8x-2y+6z=1$ and passes through the points $P_1(-1,2,5)$ and $P_2(2,1,4)$.
I can think of two methods to find the normal vector to the plane.
Method 1: Since the plane is orthogonal to $8x-2y+6z=1$, then the normal vector of the plane should be orthogonal to $(8,-2,6)$. So one normal vector would be $(1,1,-1)$.
Method 2: Find the cross product of $(8,-2,6)$ and $\mathbf{P}_1\mathbf{P_2}$.
I think there is something wrong with method 1 but I can not spot it.
Could you please clarify this for me?
Best Answer
The method (2) is correct since the normal vector to the searched plane have to be orthogonal to the vector $P_1-P_2= (-3,1,1)$ and to the normal vector to the given plane $ (8,-2,6)$.
For the method (1) note that an orthogonal vector to $ (8,-2,6)$ has components such that: $(8,-2,6)(x,y,z)^T=0$ so you can fix only one component from this equation and the other two can be fixed imposing that the plane pass through the two given points.