Plane equation expressed in new orthonormal basis Let $\pi:x+y+z=0$ describe a plane in the standardbasis. $\pi$ expressed in the new basis $b_1,b_2,b_3$ is $\sqrt{2}\hat{x}+\hat{y}=0$. What is the new basis?
I have read a chapter on change of basis but I honestly have no idea what to do here. Maybe a solution and a link to a text that treats these kind of question would help.
 A: Let $B$ be the standard basis and $B’$ be the new basis.
I would now use the notation $[\mathbf{v}]_{B}$ to represent the vector $\mathbf{v}$ expressed in the basis $B$.
We note from the two plane equations that the normal vector of the plane in the bases $B$ and $B’$ are $\begin{bmatrix}1 \\ 1 \\ 1 \end{bmatrix}_B$ and  $\begin{bmatrix}\sqrt{2} \\ {1} \\ 0 \end{bmatrix}_{B'}$ respectively.
These two normal vectors are actually the same vector in the two different bases and can be connected by:
$$
\sqrt{2}[b_1]_B+1[b_2]_B+0[b_3]_B=\begin{bmatrix}1 \\ 1 \\ 1 \end{bmatrix}_B
$$
Let $[b_1]_B = \begin{bmatrix}a \\ d \\ g \end{bmatrix}_B$, $[b_2]_B = \begin{bmatrix}b \\ e \\ h \end{bmatrix}_B$ and $[b_3]_B = \begin{bmatrix}c \\ f \\ i \end{bmatrix}_B$. Then we can write down:
$$
\begin{bmatrix}a & b &c \\ d &e &f \\ g & h & i\end{bmatrix}
\begin{bmatrix}\sqrt{2} \\ {1} \\ 0 \end{bmatrix}_{B'}
=
\begin{bmatrix}1 \\ 1 \\ 1 \end{bmatrix}_B
$$
$$
\sqrt{2}\begin{bmatrix}a  \\ d  \\ g \end{bmatrix}+1\begin{bmatrix}b  \\ e  \\ h \end{bmatrix}
=
\begin{bmatrix}1 \\ 1 \\ 1 \end{bmatrix}
$$
Let $\begin{bmatrix}a \\ d \\ g \end{bmatrix}
=
\begin{bmatrix}{1 \over \sqrt{2}}  \\ 0 \\ {1 \over \sqrt{2}} \end{bmatrix}$, then $\begin{bmatrix}b \\ e \\ h \end{bmatrix}
=
\begin{bmatrix}0  \\ 1 \\ 0 \end{bmatrix}$ and $\begin{bmatrix}c \\ f \\ i \end{bmatrix}
=b_1\times b_2
=\begin{bmatrix}{-1 \over \sqrt{2}}  \\ 0 \\ {1 \over \sqrt{2}} \end{bmatrix}$
