Find an equation of the plane through $A,B,C$ in the form $v\vec n=d$.

Question

Let $A,B,C$ be points in $\mathbb R^3$ with position vectors $a=(1,2,-3), b=(2,1,1), c=(-3,4,-4)$. I know how to find $n$ and $d$. But I don't get how to show it in the form $v \vec n=d$. What are the other possible forms of the equation of a plane apart from this one and Cartesian ?

Answer 1

Given a fixed point $A$ in the plane, and a normal vector $\vec{n}$, if a point $P$ is in the plane, the vector $\vec{AP}$ must be orthogonal to $\vec{n}$, that is, $\vec{AP} \cdot \vec{n} = 0$. Remember that we usually write $\vec{AP} = P - A$. That would give: $(P - A) \cdot \vec{n} = 0$, and then $P \cdot \vec{n} = A \cdot \vec{n}$. Notice that $A$ is fixed, so $A \cdot \vec{n}$ is a constant. I am abusing heavily on the notation here, because we don't do dot products with points, but we can interprete $A$ as $\vec{OA}$, where $O$ is the origin of the Cartesian system of coordinates. Then, your $d$ would be $A \cdot \vec{n}$, and $v$ would be $P \cdot \vec{n}$.