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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.

John Keeper
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4 Answers4

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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.

A.Γ.
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  • Then the normal vector of the plane with the new basis is $(0,0,1)$. Now a point that belongs to the plane in the original basis is $(0,0,1/2)$. This point in the new basis is $P^{-1}(0,0,1/2)$ where $P$ is the change of basis matrix. Therefore the equation of the plane in the new basis is $0(x'-(\sqrt{6}/6)+0(y'-0)+1(z'+\sqrt{3}/6)=0$. So $z'=-\sqrt{3}/6$ Is the equation of the plane in the new basis.($(\sqrt{6}/6,0,\sqrt{3}/6)$ is the point that belongs to the plane with the new basis). Is this correct? – John Keeper Oct 13 '18 at 15:11
  • @JohnKeeper The normal vector is correct (easy to see as the third basic vector is the normal vector). The new coordinates for the point $(0,0,1/2)$ is though $(0,-\frac{1}{2\sqrt{3}},\frac{1}{\sqrt{6}})$. – A.Γ. Oct 13 '18 at 17:04
  • @JohnKeeper Note that you have to fix a particular order of the basic vectors before getting the transformation matrix. – A.Γ. Oct 13 '18 at 18:52
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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 $$\begin{bmatrix} x\\y\\z \end{bmatrix}=P\begin{bmatrix} X\\Y\\Z \end{bmatrix}. $$

Bernard
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  • Your $P$ is a permutation of the transformation matrix. The first column should be the coordinates of the first new basic vector etc. – A.Γ. Oct 13 '18 at 17:08
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    @A.Γ.: When I posted, it was not a permutation of the transformation matrix. It seems the order of the vectors in the new basis was changed in the edit which happened after my post. I'll fix that in a moment. – Bernard Oct 13 '18 at 18:38
  • Right, I see it now that the post has been edited. – A.Γ. Oct 13 '18 at 18:49
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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.

amd
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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?

Dr. Sonnhard Graubner
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