State and explain the theorem of perpendicular axes digram
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ans..Given: Ellipsoid {\displaystyle \ {\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{b^{2}}}+{\frac {z^{2}}{c^{2}}}=1\ } and the plane with equation {\displaystyle \ n_{x}x+n_{y}y+n_{z}z=d\ ,} which have an ellipse in common.
Wanted: Three vectors {\displaystyle {\vec {f}}_{0}} (center) and {\displaystyle {\vec {f}}_{1},\;{\vec {f}}_{2}} (conjugate vectors), such that the ellipse can be represented by the parametric equation
{\displaystyle {\vec {x}}={\vec {f}}_{0}+{\vec {f}}_{1}\cos t+{\vec {f}}_{2}\sin t\quad } (See ellipse).

Plane section of the unit sphere (See example)
Solution: The scaling {\displaystyle \ u={\frac {x}{a}}\ ,\ v={\frac {y}{b}}\,\ w={\frac {z}{c}}\ } transforms the ellipsoid onto the unit sphere {\displaystyle u^{2}+v^{2}+w^{2}=1\ } and the given plane onto the plane with equation {\displaystyle \ n_{x}au+n_{y}bv+n_{z}cw=d\ }. Let {\displaystyle \ m_{u}u+m_{v}v+m_{w}w=\delta \ } be the HESSE normal form of the new plane and {\displaystyle \;{\vec {m}}=(m_{u},m_{v},m_{w})^{T}\;} its unit normal vector. Hence
{\displaystyle {\vec {e}}_{0}=\delta \;{\vec {m}}\;} is the center of the intersection circle and {\displaystyle \;\rho ={\sqrt {1-\delta ^{2}}}\;} its radius (See diagram).
In case of {\displaystyle \ m_{w}=\pm 1\ } let be {\displaystyle \quad {\vec {e}}_{1}=(\rho ,0,0)^{T}\;,\ {\vec {e}}_{2}=(0,\rho ,0)^{T}\;.} (The plane is horizontal !)
In case of {\displaystyle \ m_{w}\neq \pm 1\ } let be {\displaystyle \quad {\vec {e}}_{1}=\rho \,{\frac {(m_{v},-m_{u},0)^{T}}{\sqrt {m_{u}^{2}+m_{v}^{2}}}}\;,\ {\vec {e}}_{2}={\vec {m}}\times {\vec {e}}_{1}\ .}
In any case the vectors {\displaystyle {\vec {e}}_{1},{\vec {e}}_{2}} are orthogonal, parallel to the intersection plane and have length {\displaystyle \rho } (radius of the circle). Hence the intersection circle can be described by the parametric equation {\displaystyle \;{\vec {u}}={\vec {e}}_{0}+{\vec {e}}_{1}\cos t+{\vec {e}}_{2}\sin t\;.}
The reverse scaling (See above) transforms the unit sphere back to the ellipsoid and the vectors {\displaystyle {\vec {e}}_{0},{\vec {e}}_{1},{\vec {e}}_{2}} are mapped onto vectors {\displaystyle {\vec {f}}_{0},{\vec {f}}_{1},{\vec {f}}_{2}}, which were wanted for the parametric representation of the intersection ellipse.
How to find the vertices and semi-axes of the ellipse is described in ellipse.
Example: The diagrams show an ellipsoid with the semi-axes {\displaystyle \;a=4,\;b=5,\;c=3\;} which is cut by the plane {\displaystyle \;x+y+z=5\;.}
Wanted: Three vectors {\displaystyle {\vec {f}}_{0}} (center) and {\displaystyle {\vec {f}}_{1},\;{\vec {f}}_{2}} (conjugate vectors), such that the ellipse can be represented by the parametric equation
{\displaystyle {\vec {x}}={\vec {f}}_{0}+{\vec {f}}_{1}\cos t+{\vec {f}}_{2}\sin t\quad } (See ellipse).

Plane section of the unit sphere (See example)
Solution: The scaling {\displaystyle \ u={\frac {x}{a}}\ ,\ v={\frac {y}{b}}\,\ w={\frac {z}{c}}\ } transforms the ellipsoid onto the unit sphere {\displaystyle u^{2}+v^{2}+w^{2}=1\ } and the given plane onto the plane with equation {\displaystyle \ n_{x}au+n_{y}bv+n_{z}cw=d\ }. Let {\displaystyle \ m_{u}u+m_{v}v+m_{w}w=\delta \ } be the HESSE normal form of the new plane and {\displaystyle \;{\vec {m}}=(m_{u},m_{v},m_{w})^{T}\;} its unit normal vector. Hence
{\displaystyle {\vec {e}}_{0}=\delta \;{\vec {m}}\;} is the center of the intersection circle and {\displaystyle \;\rho ={\sqrt {1-\delta ^{2}}}\;} its radius (See diagram).
In case of {\displaystyle \ m_{w}=\pm 1\ } let be {\displaystyle \quad {\vec {e}}_{1}=(\rho ,0,0)^{T}\;,\ {\vec {e}}_{2}=(0,\rho ,0)^{T}\;.} (The plane is horizontal !)
In case of {\displaystyle \ m_{w}\neq \pm 1\ } let be {\displaystyle \quad {\vec {e}}_{1}=\rho \,{\frac {(m_{v},-m_{u},0)^{T}}{\sqrt {m_{u}^{2}+m_{v}^{2}}}}\;,\ {\vec {e}}_{2}={\vec {m}}\times {\vec {e}}_{1}\ .}
In any case the vectors {\displaystyle {\vec {e}}_{1},{\vec {e}}_{2}} are orthogonal, parallel to the intersection plane and have length {\displaystyle \rho } (radius of the circle). Hence the intersection circle can be described by the parametric equation {\displaystyle \;{\vec {u}}={\vec {e}}_{0}+{\vec {e}}_{1}\cos t+{\vec {e}}_{2}\sin t\;.}
The reverse scaling (See above) transforms the unit sphere back to the ellipsoid and the vectors {\displaystyle {\vec {e}}_{0},{\vec {e}}_{1},{\vec {e}}_{2}} are mapped onto vectors {\displaystyle {\vec {f}}_{0},{\vec {f}}_{1},{\vec {f}}_{2}}, which were wanted for the parametric representation of the intersection ellipse.
How to find the vertices and semi-axes of the ellipse is described in ellipse.
Example: The diagrams show an ellipsoid with the semi-axes {\displaystyle \;a=4,\;b=5,\;c=3\;} which is cut by the plane {\displaystyle \;x+y+z=5\;.}
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