Tilted Ellipsoid

The equation for the initial ellipse is

$r^t A r = 1$

where $r = [x, y, z]^t$ and matrix A is given by

$A = \begin{bmatrix} 1 / a^2 & 0 & 0 \\ 0 & 1 / b^2 & 0 \\ 0 & 0 & 1 / c^2 \end{bmatrix}$

The consecutive rotations about the x' axis and y' axis and z' axes, results in the following medication to the above equation:

$r^t T A T^t r = 1$

where $T = R_x(\theta_x) R_y(\theta_y)R_z(\theta_z)$

with $\theta_x = -60^{\circ}$ and $\theta_y = -45^{\circ}$ and $\theta_z = -30^{\circ}$ .

$R_x$ , $R_y$ and $R_z$ are the standard rotation matrices about the x-, y- and z-axes.

Now, the normal to the surface of the ellipsoid is given by the gradient, and this gradient must be a positive multiple of the unit vector along the z-axis, which is unit vector $\hat {k}$

$T A T^t r = c \hat {k} , c > 0$

It follows that

$r = c T A^{-1} T^t \hat {k}$

substituting this into the equation of the ellipse, results in

$c = { 1 \over \sqrt { {\hat {k}}^t T A^{-1} T^t \hat {k} }}$

therefore, the peak point is given by

$r ={ T A^{-1} T^t \hat {k} \over \sqrt { {\hat {k}}^t T A^{-1} T^t \hat {k} }}$

and the z-coordinate of this peak point is given by

$z = { \hat {k} }^t r$

which upon simplification becomes

$z = \sqrt { {\hat {k}}^t T A^{-1} T^t \hat {k} }$

To find the numerical value of z , we have to evaluate matrix T numerically and substitute in the above expression.