Show that point does not belong to a plane

Question

Problem

Show that point $\textbf{q}$ does not belong to plane defined by these 3 points:

$$ \textbf{p}_1 = \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix}, \textbf{p}_2=\begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix}, \textbf{p}_3 = \begin{bmatrix} 2 \\ 4 \\ 0 \end{bmatrix} $$

Point $\textbf{q}$ is defined as.

$$ \textbf{q}=\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} $$

Attempt to solve

Plane in vector form can be defined as:

$$ \textbf{x} = \textbf{p} + s\textbf{u}+ t\textbf{v}, \{s,t\} \in \mathbb{R} $$

Where vectors $\textbf{u}$ and $\textbf{v} $ define the plane and $\textbf{p}$ is point which belongs to the plane. let $\textbf{p} = \textbf{p}_2$ and vectors $\textbf{u}$ and $\textbf{v}$ can be defined as

$$ \textbf{u} = \textbf{p}_2 - \textbf{p}_1, \textbf{v} = \textbf{p}_2 - \textbf{p}_3 $$

$$ \textbf{u} = \begin{bmatrix} 0 - 2 \\ 1 -1 \\ -1 -1 \end{bmatrix}, \textbf{v} = \begin{bmatrix} 0 - 2 \\ 1 - 4 \\ -1 - 0 \end{bmatrix} $$

$$ \textbf{u} = \begin{bmatrix} -2 \\ 0 \\ -2 \end{bmatrix}, \textbf{v} = \begin{bmatrix} -2 \\ -3 \\ -1 \end{bmatrix} $$

Now we can write the equation as:

$$ \textbf{x}= \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} + s \begin{bmatrix} -2 \\ 0 \\ -2 \end{bmatrix} + t \begin{bmatrix} -2 \\ -3 \\ -1 \end{bmatrix} $$

Now i want to show that $$ \nexists \{s,t\} : \textbf{p} + s\text{u} + t \textbf{v} = \begin{bmatrix} 1 \\ 1 \\1 \end{bmatrix} $$

I can form matrix equation:

$$\begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix} + s \begin{bmatrix} -2 \\ 0 \\ -2 \end{bmatrix} + t \begin{bmatrix} -2 \\ -3 \\ -1 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} $$

$$ \implies s\begin{bmatrix} -2 \\ 0 \\ -2 \end{bmatrix} + t \begin{bmatrix} -2 \\ -3 \\ -1 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}-\begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix} $$

$$ \implies s\begin{bmatrix} -2 \\ 0 \\ -2 \end{bmatrix} + t \begin{bmatrix} -2 \\ -3 \\ -1 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix} $$

$$ \begin{bmatrix} -2 & -2 \\ 0 & -3 \\ -2 & -1 \end{bmatrix} \begin{bmatrix} s \\ t \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix} $$

Now i can solve the equation with row redcution

$$ \begin{bmatrix} -2 & -2 & 1 \\ 0 & -3 & 0 \\ -2 & -1 & 2 \end{bmatrix} $$

Solution is $$ s = -\frac{15}{19}, t = \frac{1}{19} $$

Problem is solution should be impossible. i Was trying to prove poin $\textbf{q}$ doe $\textbf{not}$ belong to plane. Is my approach correct or did something went wrong with my solution ?

Answer 1

Let me check your solution:

$$(-2) \left( -\frac{15}{19}\right)+(-2) \left( \frac{1}{19}\right)= \left( \frac{2}{19}\right)(15-1) \ne 1$$

Hence you made a mistake in solving for your $s$ and $t$.

Here is the RREF, we can see that the augmented matrix is not consistent.

octave:1> rref([-2 -2 1; 0 -3 0 ; -2 -1 2])
ans =

   1   0   0
   0   1   0
   0   0   1

Answer 2

Define ${\bf u} = {\bf p}_2 - {\bf p}_1$ and ${\bf v} = {\bf p}_3 - {\bf p}_1$, a normal vector to the plane is

$$ \hat{\bf n} = \frac{{\bf u}\times {\bf v}}{|{\bf u}\times {\bf v}|} = \frac{1}{\sqrt{19}}\pmatrix{3 \\ -1 \\ -3} $$

A point ${\bf x}$ belongs to the plane if

$$ \hat{\bf n}\cdot({\bf x} - {\bf p}_1) = 0 $$

If you set ${\bf x} = {\bf q}$ you'll find

$$ \hat{\bf n}\cdot({\bf q} - {\bf p}_1) = -\frac{3}{\sqrt{19}} \not=0 $$

So the point ${\bf q}$ does not belong to the plane