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?
$\endgroup$ 51 Answer
$\begingroup$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.
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