How can we derive $ax + by + cz + d = 0$ from a point and two non-colinear vectors using mixed product?

Question

I'm trying to figure this out... The mixed product gives us the volume of the parallelepiped which the three vectors form. However, I don't see the connection between a point (let's call it $T_0(x_0, y_0, z_0)$) and two non-colinear vectors (let's name them $\vec a$ and $\vec b$.

Can someone guide me/give me some hints?

Answer 1

I assume that we are talking about a plane in $\mathbb{R}^3$. Observe that each point $(x,y,z)$ on the plane can be characterized by the following property: the vector that points from $T_0$ to $(x,y,z)$ and the two non-colinear vectors than span the plane form a parallelepiped of volume $0$. If you put this in your formula for the mixed product, you will get something that looks like $ax+by+cz+d=0$.