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[SOLVED] Equation of plane

dwsmith

Well-known member
Feb 1, 2012
1,673
Equation of a plane after a coordinate transformation. Not sure about the second part in regards to finding the plane in the new system.
The angles between the respective axes $O_{x_1'x_2'x_3'}$ and the $O_{x_1x_2x_3}$ Cartesian system are given by the table below

\[x_1\]\[x_2\]\[x_3\]
\[x'_1\]\[\frac{\pi}{4}\]\[\frac{\pi}{2}\]\[\frac{\pi}{4}\]
\[x'_2\]\[\frac{\pi}{3}\]\[\frac{\pi}{4}\]\[\frac{2\pi}{3}\]
\[x'_3\]\[\frac{2\pi}{3}\]\[\frac{\pi}{4}\]\[\frac{\pi}{3}\]

Determine the transformation matrix between the two sets of axes
$$
[A] = \begin{bmatrix}
\frac{\sqrt{2}}{2} & 0 & \frac{\sqrt{2}}{2}\\
\frac{1}{2} & \frac{\sqrt{2}}{2} & -\frac{1}{2}\\
-\frac{1}{2} & \frac{\sqrt{2}}{2} & \frac{1}{2}
\end{bmatrix}
$$
The matrix $[A]$ is the transformation matrix from the new coordinate system to the old.
The equation of the plane $x_1 + x_2 + x_3 = \frac{1}{\sqrt{2}}$ in its primed axes form, that is, in the form $b_1x_1' + b_2x_2' +b_3x_3' = b$.
\begin{alignat*}{3}
\begin{bmatrix}
x_1'\\
x_2'\\
x_3'
\end{bmatrix} & = &
\begin{bmatrix}
\frac{\sqrt{2}}{2} & 0 & \frac{\sqrt{2}}{2}\\
\frac{1}{2} & \frac{\sqrt{2}}{2} & -\frac{1}{2}\\
-\frac{1}{2} & \frac{\sqrt{2}}{2} & \frac{1}{2}
\end{bmatrix}
\begin{bmatrix}
x_1\\
x_2\\
x_3
\end{bmatrix}
\end{alignat*}
 
Last edited by a moderator:

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
The matrix $[A]$ is the transformation matrix from the new coordinate system to the old.
The equation of the plane $x_1 + x_2 + x_3 = \frac{1}{\sqrt{2}}$ in its primed axes form, that is, in the form $b_1x_1' + b_2x_2' +b_3x_3' = b$.
\begin{alignat*}{3}
\begin{bmatrix}
x_1'\\
x_2'\\
x_3'
\end{bmatrix} & = &
\begin{bmatrix}
\frac{\sqrt{2}}{2} & 0 & \frac{\sqrt{2}}{2}\\
\frac{1}{2} & \frac{\sqrt{2}}{2} & -\frac{1}{2}\\
-\frac{1}{2} & \frac{\sqrt{2}}{2} & \frac{1}{2}
\end{bmatrix}
\begin{bmatrix}
x_1\\
x_2\\
x_3
\end{bmatrix}
\end{alignat*}
The matrix $A$ is orthogonal, that is $A^{t}=A^{-1}$, so $X'=AX$ is equivalent to $X=A^tX'$. In our case,
$$\begin{bmatrix}{x_1}\\{x_2}\\{x_3}\end{bmatrix}=\begin{bmatrix}{\sqrt{2}/2}&{1/2}&{-1/2}\\{0}&{\sqrt{2}/2}&{\sqrt{2}/2}\\{\sqrt{2}/2}&{-1/2}&{1/2}\end{bmatrix}\begin{bmatrix}{x'_1}\\{x'_2}\\{x'_3}\end{bmatrix}$$
Now, $x_1+x_2+x_3=\dfrac{1}{\sqrt{2}}\Leftrightarrow \left(\dfrac{\sqrt{2}}{2}x'_1+\dfrac{1}{2}x'_2-\dfrac{1}{2}x'_3\right)+\ldots=\dfrac{1}{\sqrt{2}}$ and we get the equation of the plane in its primed axes form.