Ytukyg

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:

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

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 $$ nexists {s,t} : textbf{p} + stext{u} + t textbf{v} = begin{bmatrix} 1 \ 1 \1 end{bmatrix} $$

I can form matrix equation:

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

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

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

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

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 ?

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

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