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I'm reviewing for the math GRE. Here's a linear algebra question from an old text.

"Find an equation for the plane that is perpendicular to the plane $$8x-2y+6z=1$$ and passes through the points $P_1(-1,2,5)$ and $P_2(2,1,4).$"

Flawed Solution:

Since our plane is perpendicular to a plane with normal vector $\langle8,-2,6\rangle,$ our plane's normal vector is a scalar multiple of the vector $\langle 1,1,-1\rangle$ (since $\langle 8,-2,6\rangle \cdot \langle1,1,-1\rangle=0$).

Since we have a normal vector and a point $P_1$, our plane is uniquely determined: Given any point $(x,y,z)$ on our plane, the equation of our plane is $$\langle 1,1,-1\rangle \cdot\langle x+1,y-2,z-5\rangle =0$$

Simplifying we get $$x+y-z+4=0$$

However, this is the wrong plane since $P_2$ is not on this plane. Where did I go wrong?

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up vote 1 down vote accepted

Your solution is wrong. Vector $(1, 1, -1)$ is perpendicular to vector $(8, -2, 6)$ but it is not a plane' s normal vector. The plane' s normal vector is the vector product of $(8, -2, 6)$ and vector $P_1P_2=(2, 1, 4)-(-1, 2, 5)=(3, -1, -1)$.

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There are many possible normals to a plane perpendicular to a given plane. Think about the first plane being $x=0$ with normal vector $\langle1,0,0\rangle$The normal vector for a perpendicular plane could be $\langle0,1,0\rangle$ but it could also be $\langle0,0,1\rangle$ or any linear combination of those.

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