# The equation of a plane perpendicular to another plane

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?

• In method 1, you find the normal of the vector and then you need to find $d$.You need only one point to do that (i.e $P1$ or $P2$), so if they gave you two points, then there is probably something wrong there. (Usually, there is no useless or unused data in math questions...) Jun 5, 2015 at 11:32
• @Eminem there are an infinite number of planes through one point which are perpendicular to a given plane (rotate about the point with axis normal to the plane).
– Paul
Jun 5, 2015 at 11:42
• Method 1 gives you the direction of a line in your original plane. There is no reason why this direction should be orthogonal to the plane you require. Method 2 will work though as you now have 2 directions in the plane you have to find.
– Paul
Jun 5, 2015 at 11:49
• @Paul But if you have the normal of the plane, you have to find $d$ to complete the task. So here, if the normal is $(1,1,-1)$ than the equation is $x+y-z+d=0$. Then you put $P1$ or $P2$ there to find $d$, and that is it, right? Jun 5, 2015 at 14:19
• @Eminem (1, 1, -1) is just one direction in the original plane. It is a normal to some plane alright, but there is no reason to suppose that it is normal to the plane through $P_1$ and $P_2$. Infact, it can't be, as those two points give different values for d.
– Paul
Jun 5, 2015 at 18:51

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.