I want to check if the four points of $\mathbb{R}^3$ $$P=(1,2,2), \ Q=(3,5,6), \ R=(1,3,2), \ S=(5,2,3)$$ lie on one plane.
For that do we have to take three points, to define the plane and check if the forth point is on that plane?
For that do we do the following ?
We consider the points $P,Q,R$.
$\vec{n}$, the normal vector to the plane is defined as $\vec{n}=\vec{PQ}\times \vec{PR}$, i.e. we have $\vec{n}=(2,3,4)\times(0,1,0)=(-4, 0, 2)$.
The equation of the plane is \begin{align*}n_x(x-P_x)+n_y(y-P_y)+n_z(z-P_z)=0 &\Rightarrow -4\cdot (x-1)+0\cdot (y-2)+2\cdot (z-2)=0 \\ & \Rightarrow -4x+4+2z-4=0 \\ & \Rightarrow z=2x\end{align*} The point $S$ doesn't satisfy this equation, that means that te four points don't lie on the same plane.
Is that correct ?
Best Answer
Yes, that is correct. Equivalently, you may check directly that $$\begin{align}\det\left(\overrightarrow{PQ},\overrightarrow{PR},\overrightarrow{PS}\right) &=\begin{vmatrix}2&0&4\\3&1&0\\4&0&1\end{vmatrix}\\ &=\begin{vmatrix}2&4\\4&1\end{vmatrix}\\ &=-14\ne0.\end{align}$$