Given are these points $$v_1=\begin{pmatrix} 1\\ 2\\ 3
\end{pmatrix}, v_2=\begin{pmatrix} 2\\ 3\\ 3 \end{pmatrix},
v_3=\begin{pmatrix} 2\\ 4\\ 4 \end{pmatrix}$$Determine the normal to the plane which contains these 3 points.
My problem with this is I'm not sure what is really meant with "normal". Is it just the normal-vector or I will also need some parameter multiplied with the normal-vector?
So far I've done that but I'm not sure if that is all what is asked?
First calculate both direction vectors:
$$\begin{pmatrix}
2\\
3\\
3
\end{pmatrix}-
\begin{pmatrix}
1\\
2\\
3
\end{pmatrix}=
\begin{pmatrix}
1\\
1\\
0
\end{pmatrix} \text{ and }
\begin{pmatrix}
2\\
4\\
4
\end{pmatrix}-
\begin{pmatrix}
1\\
2\\
3
\end{pmatrix}=
\begin{pmatrix}
1\\
2\\
1
\end{pmatrix}$$
Now cross-product of these two vectors gives the normal vector:
$$\begin{pmatrix}
1\\
1\\
0
\end{pmatrix}
\times
\begin{pmatrix}
1\\
2\\
1
\end{pmatrix}=
\begin{pmatrix}
1\\
-1\\
1
\end{pmatrix}$$
But now is this really an answer to the question? Or I need some parametric form of it? Please help :/
Best Answer
“The” normal to a plane typically really does refer to any vector that is normal to the plane. In other words, your suggested answer (or any non-zero scalar multiple of it) is perfectly fine.
No need to express the normal generally as $$k\begin{pmatrix} 1\\ -1\\ 1 \end{pmatrix}\quad(k\in\mathbb R\setminus\{0\}).$$