Editing Appendix/Ramblings/ConcentricEllipsoidalT8Coordinates (section)

=Angle Between Unit Vectors=
We begin by restating that the coordinate, scale factor, and unit vector associated with the normal to our ellipsoidal surface are,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\lambda_1</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~(x^2 + q^2y^2 + p^2z^2)^{1 / 2} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~h_1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\lambda_1 \ell_{3D} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\hat{e}_1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\hat\imath (x\ell_{3D} )
+
\hat\jmath (q^2 y\ell_{3D} )
+
\hat{k} (p^2z\ell_{3D} ) \, .
</math>
  </td>
</tr>
</table>

In the [[#TableKappa8|Table below titled, "Direction Cosines Components for &kappa;8 Coordinates"]], there are two fully-formed unit vectors that are each orthogonal to the (first) unit vector that is normal to the ellipsoid's surface.  Here we will refer to the coordinates of these two fully-formed unit vectors as,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\lambda_2</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{xy^{1/q^2}}{z^{2/p^2}}</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp; </td>
  <td align="right">
<math>~\kappa_2</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{(qy)^{1/q^2}}{x} \, .</math>
  </td>
</tr>
</table>
The associated scale factors and unit vectors are given by the following expressions:

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~h_{\lambda_2}^{-2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\biggl(\frac{\partial\lambda_2}{\partial x} \biggr)^2
+
\biggl(\frac{\partial\lambda_2}{\partial y} \biggr)^2
+
\biggl(\frac{\partial\lambda_2}{\partial z} \biggr)^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\biggl(\frac{\lambda_2}{x} \biggr)^2
+
\biggl(\frac{\lambda_2}{q^2y} \biggr)^2
+
\biggl(- \frac{2\lambda_2}{p^2z} \biggr)^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \lambda_2^2 \biggl[
\frac{q^4y^2p^4z^2 + x^2p^4z^2 + 4x^2 q^4y^2}{x^2 q^4y^2 p^4z^2} 
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ h_{\lambda_2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\frac{x q^2y p^2z}{\lambda_2 \mathcal{D}}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \hat{e}_{\lambda_2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\hat\imath \biggl[ h_{\lambda_2} \biggl(\frac{\partial \lambda_2}{\partial x}\biggr)\biggr]
+
\hat\jmath \biggl[ h_{\lambda_2} \biggl(\frac{\partial \lambda_2}{\partial y}\biggr)\biggr]
+
\hat{k} \biggl[ h_{\lambda_2} \biggl(\frac{\partial \lambda_2}{\partial z}\biggr)\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\hat\imath \biggl[ \frac{q^2y p^2z}{\mathcal{D}} \biggr]
+
\hat\jmath \biggl[ \frac{xp^2z}{\mathcal{D}} \biggr]
-
\hat{k} \biggl[ \frac{2x q^2y }{\mathcal{D}} \biggr] \, .
</math>
  </td>
</tr>
</table>


<table border="1" width="80%" align="center" cellpadding="8"><tr><td align="left">
With regard to orthogonality, note that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\hat{e}_1 \cdot \hat{e}_{\lambda_2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[
\hat\imath (x\ell_{3D} )
+
\hat\jmath (q^2 y\ell_{3D} )
+
\hat{k} (p^2z\ell_{3D} ) 
\biggr] \cdot
\biggl\{
\hat\imath \biggl[ \frac{q^2y p^2z}{\mathcal{D}} \biggr]
+
\hat\jmath \biggl[ \frac{xp^2z}{\mathcal{D}} \biggr]
-
\hat{k} \biggl[ \frac{2x q^2y }{\mathcal{D}} \biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(x\ell_{3D} )\biggl[ \frac{q^2y p^2z}{\mathcal{D}} \biggr]
+
(q^2 y\ell_{3D} )\biggl[ \frac{xp^2z}{\mathcal{D}} \biggr]
-
(p^2z\ell_{3D} )\biggl[ \frac{2x q^2y }{\mathcal{D}} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
0 \, .</math>
  </td>
</tr>
</table>

</td></tr></table>


And,

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~h_{\kappa_2}^{-2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\biggl(\frac{\partial\kappa_2}{\partial x} \biggr)^2
+
\biggl(\frac{\partial\kappa_2}{\partial y} \biggr)^2
+
\biggl(\frac{\partial\kappa_2}{\partial z} \biggr)^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\biggl(-\frac{\kappa_2}{x} \biggr)^2
+
\biggl(\frac{\kappa_2}{q^2 y} \biggr)^2
=
\frac{\kappa_2^2}{x^2q^4y^2}\biggl[x^2 + q^4y^2\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~h_{\kappa_2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\frac{xq^2y}{\kappa_2 (x^2 + q^4y^2)^{1 / 2}} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~\hat{e}_{\kappa_2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
-~\hat\imath \biggl[ \frac{q^2y}{(x^2 + q^4y^2)^{1 / 2}}\biggr]
+
\hat\jmath \biggl[ \frac{x}{(x^2 + q^4y^2)^{1 / 2}} \biggr]
+
\hat{k} \biggl[ 0 \biggr] \, .
</math>
  </td>
</tr>
</table>


<table border="1" width="80%" align="center" cellpadding="8"><tr><td align="left">
Again, note that with regard to orthogonality, 
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\hat{e}_1 \cdot \hat{e}_{\kappa_2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[
\hat\imath (x\ell_{3D} )
+
\hat\jmath (q^2 y\ell_{3D} )
+
\hat{k} (p^2z\ell_{3D} ) 
\biggr] \cdot
\biggl\{
-\hat\imath \biggl[ \frac{q^2y}{(x^2 + q^4y^2)^{1 / 2}}\biggr]
+
\hat\jmath \biggl[ \frac{x}{(x^2 + q^4y^2)^{1 / 2}} \biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-~(x\ell_{3D} )\biggl[ \frac{q^2y}{(x^2 + q^4y^2)^{1 / 2}} \biggr]
+
(q^2 y\ell_{3D} )\biggl[ \frac{x}{(x^2 + q^4y^2)^{1 / 2}}  \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
0 \, .</math>
  </td>
</tr>
</table>

</td></tr></table>


From this pair of orthogonality checks, we appreciate that both unit vectors always lie in the plane that is tangent to the surface of our ellipsoid.  Next, let's determine the angle, <math>~\alpha</math>, between these two unit vectors as measured in the relevant tangent-plane.

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\cos\alpha \equiv \hat{e}_{\lambda_2} \cdot \hat{e}_{\kappa_2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{
\hat\imath \biggl[ \frac{q^2y p^2z}{\mathcal{D}} \biggr]
+
\hat\jmath \biggl[ \frac{xp^2z}{\mathcal{D}} \biggr]
-
\hat{k} \biggl[ \frac{2x q^2y }{\mathcal{D}} \biggr] 
\biggr\} 
\biggl\{
-\hat\imath \biggl[ \frac{q^2y}{(x^2 + q^4y^2)^{1 / 2}}\biggr]
+
\hat\jmath \biggl[ \frac{x}{(x^2 + q^4y^2)^{1 / 2}} \biggr]
+
\hat{k} \biggl[ 0 \biggr] 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\biggl[ \frac{q^2y}{(x^2 + q^4y^2)^{1 / 2}}\biggr]\biggl[ \frac{q^2y p^2z}{\mathcal{D}} \biggr]
+
\biggl[ \frac{x}{(x^2 + q^4y^2)^{1 / 2}} \biggr]\biggl[ \frac{xp^2z}{\mathcal{D}} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{p^2 z}{\mathcal{D} (x^2 + q^4y^2)^{1 / 2}} \biggr] \biggl[ x^2 - q^4y^2 \biggr] \, .
</math>
  </td>
</tr>
</table>


----


Let's again visit the unit vector that we know lies in the tangent-plane and is always orthogonal to <math>~\lambda_2</math>, namely,

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\hat{e}_{\lambda_3}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
-~\hat\imath \biggl[ x(2q^4 y^2 + p^4z^2 )\biggr]\frac{\ell_{3D}}{\mathcal{D}}
+
\hat\jmath \biggl[q^2y(p^4z^2 + 2x^2) \biggr]\frac{\ell_{3D}}{\mathcal{D}}
+
\hat{k} \biggl[p^2z (x^2 - q^4y^2) \biggr]\frac{\ell_{3D}}{\mathcal{D}} 
\, .
</math>
  </td>
</tr>
</table>

We acknowledge that,

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\hat{e}_{\lambda_2} \cdot \hat{e}_{\lambda_3}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{
\hat\imath \biggl[ \frac{q^2y p^2z}{\mathcal{D}} \biggr]
+
\hat\jmath \biggl[ \frac{xp^2z}{\mathcal{D}} \biggr]
-
\hat{k} \biggl[ \frac{2x q^2y }{\mathcal{D}} \biggr] 
\biggr\} 
\biggl\{
-~\hat\imath \biggl[ x(2q^4 y^2 + p^4z^2 )\biggr]\frac{\ell_{3D}}{\mathcal{D}}
+
\hat\jmath \biggl[q^2y(p^4z^2 + 2x^2) \biggr]\frac{\ell_{3D}}{\mathcal{D}}
+
\hat{k} \biggl[p^2z (x^2 - q^4y^2) \biggr]\frac{\ell_{3D}}{\mathcal{D}} 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-~\biggl[ xq^2yp^2z(2q^4 y^2 + p^4z^2 )\biggr]\frac{\ell_{3D}}{\mathcal{D}^2}
+
\biggl[x q^2y p^2z(p^4z^2 + 2x^2) \biggr]\frac{\ell_{3D}}{\mathcal{D}^2}
-
\biggl[2xq^2y p^2z (x^2 - q^4y^2) \biggr]\frac{\ell_{3D}}{\mathcal{D}^2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[
-(2q^4 y^2 + p^4z^2 )
+
(p^4z^2 + 2x^2) 
-
2 (x^2 - q^4y^2)  
\biggr] \frac{(x q^2y p^2z) \ell_{3D}}{\mathcal{D}^2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
0 \, .
</math>
  </td>
</tr>
</table>