Find equation of plane from 2 reference systems

Question

Consider in $R^{3}$ two reference systems $S= (O; \vec{e_{1}},\vec{e_{2}},\vec{e_{3}})$ and $S'= (O'; \vec{e_{1}}',\vec{e_{2}}',\vec{e_{3}}')$, with $\vec{OO'}= (−2,3,9)$ in $S$, $\vec{e_{1}}'=\vec{e_{1}}+ 3\vec{e_{2}}+\vec{e_{3}}$, $\vec{e_{2}}'=-\vec{e_{1}}$, and $\vec{e_{3}}'= 2\vec{e_{1}}+ 5\vec{e_{2}}+ 7\vec{e_{3}}$. Find the equation in $S′$ of the plane whose equation is $2x−3y+z=2$.

I have no clue in how to start. What is the purpose of having $\vec{OO'}$? Do I have to describe $S'$ in terms of the first reference system $S$?

Any help would be really appreciated.

Answer 1

Hint.

Given a plane in $S$ given by

$$ (p-p_0)\cdot \vec n = 0 $$

and a change of coordinates to $S'$ given by the matrix $M$ and the translation $o\to o'$ we have

$$ (p-p_0)\cdot M^{\dagger}\cdot M^{-\dagger}\cdot \vec n = 0 $$

then

$$ p'-p_0' = M\cdot (p-p_0)\\ \vec n' = M^{-1}\vec n $$

after that, the translation in $p_0'$

Answer 2

"OO'" tells you where the origin of the new coordinate system is: (-2, 3, 9). If translating the origin were the only change (so that the x', y', and z' axes were parallel to the x, y, and z axes) (x', y', z') in the new coordinate system would be (x'- 2, y'+ 3, z'+ 9) in the old coordinate system.

The other change here is multiplication by the matrix $\begin{bmatrix}1 & 3 & 1 \\ -1 & 0 & 0 \\ 2 & 5 & 7 \end{bmatrix}$. But we want to determine x, y, z in terms of x', y', z' so we want the inverse of that matrix.