Given is the following plane in $R^4:$
I need to find the mirror across the plane. Hint: first find a normal vector to the plane.
The fact that it is in 4 dimensions disturbs me a bit. To find a normal vector, I can’t just use the cross product. Also, I don’t realy see how I can find the mirror using the normal vector. Thanks for your help.
(Btw: this is not homework. Our professor gave us old exams, but no solutions to them. That doesn’t realy help me.)
Edit: here are the 3 vectors of the plane if you can't see the picture
\begin{align} y &= \begin{bmatrix} 1 \\ 2 \\ 0 \\ 0 \end{bmatrix} \end{align}
\begin{align} y &= \begin{bmatrix} -1 \\ 1 \\ 3 \\ 0 \end{bmatrix} \end{align}
\begin{align} y &= \begin{bmatrix} 0 \\ 0\\ 0 \\ 1 \end{bmatrix} \end{align}
To find a normal vector $\vec n$, first find an equation of the plane, it will have the form $ \vec n \cdot \vec v =0$. Then use the normal vector together with a basis of the plane to form an appropriate basis of $R^4 $.
Another method to find the normal vector, solve a system of $3 $ equations made of dot products that must be $0 $.