Editing Appendix/Ramblings/ConcentricEllipsoidalT12Coordinates (section)

===Contrast===

The unit vector resulting (just derived) from our ''guess'' of the third-coordinate expression should be compared with [[#needed|the ''needed'' unit vector as described above]], namely,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~[\hat{e}_3]_\mathrm{needed}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
 <td align="left">
<math>~\mathfrak{A}^{-1 / 2} \mathfrak{B}^{-1 / 2} \biggl\{
\hat\imath \biggl[ x(cq^2y^2  - b p^2z^2) \biggr]
+
\hat\jmath \biggl[ y(ap^2z^2 - cx^2) \biggr]
+
\hat{k} \biggl[ z(bx^2 - aq^2y^2) \biggr]
\biggr\} \, .
</math>
  </td>
</tr>
</table>

At first glance, apart from the difference in signs, the three terms inside the curly braces appear to be identical in these two separate unit-vector expressions.  But they are not!  Relative to the ''needed'' expression, key components of each term are '''squared''' in the ''guessed'' expression.  Very close &hellip; but no cigar!

<table border="1" cellpadding="10" align="center" width="80%"><tr><td align="left">
<div align="center">'''ASIDE'''</div>
Note that, <math>~[\hat{e}_3]_\mathrm{needed} \cdot [\hat{e}_3]_\mathrm{needed} = 1</math> implies that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mathfrak{A} \mathfrak{B} = \biggl[ \frac{(abc)\mathfrak{L}}{\ell_{3D}} \biggr]^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ x(cq^2y^2  - b p^2z^2) \biggr]^2
+ \biggl[ y(ap^2z^2 - cx^2) \biggr]^2
+ \biggl[ z(bx^2 - aq^2y^2) \biggr]^2 \, .
</math>
  </td>
</tr>
</table>
But we also know that (see, for example, immediately below),
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mathfrak{A} \mathfrak{B}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(x^2 + q^4y^2 + p^4z^2)[a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2] \, .
</math>
  </td>
</tr>
</table>

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

What about the overall leading coefficient?  That is, does <math>~\mathfrak{A}\mathfrak{B} = \mathfrak{C}^2</math>&nbsp;&nbsp;? &nbsp;Well, given that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mathfrak{A} = \ell_{3D}^{-2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(x^2 + q^4y^2 + p^4z^2)</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp; </td>
  <td align="right">
<math>~\mathfrak{B} = (abc)^2\mathfrak{L}^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
[a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2] \, ,
</math>
  </td>
</tr>
</table>
we have,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mathfrak{A} \mathfrak{B}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(x^2 + q^4y^2 + p^4z^2)[a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
x^2 [a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2] 
+ q^4y^2 [a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2] 
+ p^4z^2[a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~x^2y^2z^2[a^2 + b^2q^4 +c^2p^4]
+ x^4 [b^2 z^2 + c^2 y^2] 
+ q^4y^4 [a^2 z^2  + c^2 x^2] 
+ p^4z^4[a^2 y^2 + b^2 x^2] 
</math>
  </td>
</tr>
</table>

On the other hand,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mathfrak{C}^2</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
x^2 (cq^2y^2 - bp^2z^2)^4 
+ y^2 (ap^2z^2 - cx^2)^4 
+ z^2 (aq^2y^2 - bx^2)^4 
 \, .
</math>
  </td>
</tr>
</table>