Find the equation of the plane that passes through the points

Question

Find the equation of the plane that passes through the points: $$P_1=(1,1,2)\\P_2=(2,3,3) \\P_3=(3,-3,3)$$

The answer writes: Let $x=\vec{P_1P_2}=\begin{bmatrix}1\\2\\1\end{bmatrix}$ and $y=\vec{P_1P_3}=\begin{bmatrix}2\\-4\\1\end{bmatrix}$ The normal vector $N$ must be orthogonal to both $X$ and $Y$. If we set $N=X\times Y=\begin{bmatrix}6\\1\\-8\end{bmatrix}$ Then $N$ will be a normal vector to the plane that passes through the given points. Using point $P_1$, we see that the equation of the plane is $$6(x-1)+(y-1)-8(z-1)=0$$

But I don't know what the answer means, especially how it calculates X×Y and get $\begin{bmatrix}6\\1\\-8\end{bmatrix}$, I really have no idea.

Answer 1

According to geometry, a plane can be uniquely determined given the normal vector of it and a point lying on it (the reason is intuitively simple). To solve this problem, a number of arguments are employed:

$A_1$: The cross product of two linearly independent vectors, is always orthogonal on them (linearly independent in the case of two vectors means that they are not any multiple of each other).

$A_2$: If two points belong to a plane, then so is the corresponding vector whose endpoints are the given points of the plane.

$A_3$: The cross product of two vectors $v_1=a\hat i+b\hat j+c\hat k$ and $v_2=d\hat i+e\hat j+f\hat k$ can be calculated as $$v_1\times v_2=(bf-ce)\hat i+(cd-af)\hat j+(ae-bd)\hat k$$

The normal vector then would be$$v=v_1\times v_2=\begin{bmatrix}1\\2\\1\end{bmatrix}\times \begin{bmatrix}2\\-4\\1\end{bmatrix}=\begin{bmatrix}2\times 1-1\times(-4)\\2\times 1-1\times 1\\1\times(-4)-2\times 2\end{bmatrix}=\begin{bmatrix}6\\1\\-8\end{bmatrix}$$therefore $$(x,y,z)\cdot v=k$$for some constant $k$. By substitution we finally obtain$$6x+y-8z=-1$$