Given the plane $-x-y+2z=1$ in the basis $\{(1,0,0),(0,1,0),(0,0,1)\}$ I want to calculate the equation of this plane respect the basis $\{\frac{1}{\sqrt2}(-1,1,0),\frac{1}{\sqrt3}(-1,-1,-1),\frac{1}{\sqrt6}(-1,-1,2)\}$.
I know that if I want to write a given vector $(a,b,c)$ in another basis, I just have to multiply the vector to the change of basis matrix, but I don't know that to do here.
On
Let $$A=\frac{1}{\sqrt{6}}(-1,-1,2),B=\frac{1}{\sqrt{2}}(-1,1,0),C=\frac{1}{\sqrt{3}}(-1,-1,1)$$ then we get $$\vec{x}=\vec{OA}+s\vec{AB}+t\vec{AC}$$
Can you get it from here?
On
Hint: it is an orthonormal change of basis. If you transform a normal vector for the plane and a vector for any point in the plane, you can make a transformed equation out of those.
On
Hint: The change of basis matrix is $$P=\begin{bmatrix} -\frac1{\sqrt 2} &-\frac1{\sqrt 3} &-\frac1{\sqrt 6}\\ \frac1{\sqrt 2} &-\frac1{\sqrt 3}&-\frac1{\sqrt 6}\\ 0 &-\frac1{\sqrt 3}&\frac2{\sqrt 6} \end{bmatrix}=-\frac1{\sqrt 6}\begin{bmatrix} \sqrt 3 & \sqrt 2 &1\\ -\sqrt 3 & \sqrt 2 &1\\ 0 & \sqrt 2 &-2 \end{bmatrix},$$ and if you denote $(X,Y,Z)$ the coordinates in the new basis, you have the relation $$\begin{bmatrix} x\\y\\z \end{bmatrix}=P\begin{bmatrix} X\\Y\\Z \end{bmatrix}. $$
Let $B$ be the change-of-basis matrix. Write the equation of the plane as $\mathbf n^T\mathbf v=1$, where $\mathbf n = (-1,-1,2)^T$ is the vector of coefficients in the equation. This vector is normal to the plane. Then from $\mathbf v'=B\mathbf v$ we get $$\mathbf n^T\mathbf v = \mathbf n^T(B^{-1}\mathbf v') = (\mathbf n^TB^{-1})\mathbf v' = (B^{-T}\mathbf n)^T\mathbf v' = 1.$$ In other words, if vectors transform as $\mathbf v'=B\mathbf v$, then plane normals transform as $\mathbf n'=B^{-T}\mathbf n$. In this problem, $B$ is orthogonal, so $B^{-T}=B$.
Observe that $\mathbf n$ is a scalar multiple of the third new basis vector, so you know that $\mathbf n'$ will be a corresponding multiple of $(0,0,1)^T$ without having to go through the full matrix multiplication.