# Concentric Ellipsoidal (T12) Coordinates

## Background

Building on our general introduction to Direction Cosines in the context of orthogonal curvilinear coordinate systems, and on our previous development of T3 (concentric oblate-spheroidal) and T5 (concentric elliptic) coordinate systems, here we explore the creation of a concentric ellipsoidal (T8) coordinate system. This is motivated by our desire to construct a fully analytically prescribable model of a nonuniform-density ellipsoidal configuration that is an analog to Riemann S-Type ellipsoids.

Note that, in a separate but closely related discussion, we made attempts to define this coordinate system, numbering the trials up through "T7." In this "T7" effort, we were able to define a set of three, mutually orthogonal unit vectors that should work to define a fully three-dimensional, concentric ellipsoidal coordinate system. But we were unable to figure out what coordinate function, $~\lambda_3(x, y, z)$, was associated with the third unit vector. In addition, we found the $~\lambda_2$ coordinate to be rather strange in that it was not oriented in a manner that resembled the classic spherical coordinate system. Here we begin by redefining the $~\lambda_2$ coordinate such that its associated $~\hat{e}_3$ unit vector lies parallel to the x-y plane.

The 1st coordinate and its associated unit vector are as follows:

 $~\lambda_1$ $~\equiv$ $~ (x^2 + q^2 y^2 + p^2 z^2)^{1 / 2} \, ;$ $~\hat{e}_1$ $~=$ $~ \ell_{3D} \biggl[ \hat\imath (x) + \hat\jmath (q^2y ) + \hat{k} (p^2 z) \biggr] \, ,$

where,

 $~\ell_{3D}$ $~\equiv$ $~ (x^2 + q^4y^2 + p^4 z^2)^{- 1 / 2} \, .$

## Generalized Prescription for 2nd Coordinate

### Default

Let's adopt the following generalized prescription for the 2nd coordinate:

 $~\lambda_2$ $~\equiv$ $~ x^a y^b z^c \, ,$

in which case,

 $~\hat{e}_2$ $~=$ $~ \frac{1}{\mathfrak{L}} \biggl[ \hat\imath \biggl(\frac{yz}{bc}\biggr) + \hat\jmath \biggl(\frac{xz}{ac}\biggr) + \hat{k} \biggl(\frac{xy}{ab}\biggr) \biggr] \, ,$

where,

 $~\mathfrak{L}^2$ $~\equiv$ $~ \frac{1}{a^2b^2c^2} \biggl[ a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2 \biggr] \, .$

Now, to ensure that $~\hat{e}_2$ is perpendicular to $~\hat{e}_1$, we need,

 $~\hat{e}_1 \cdot \hat{e}_2$ $~=$ $~0$ $~\Rightarrow~~~ 0$ $~=$ $~ \frac{\ell_{3D}}{\mathfrak{L}} \biggl[ \frac{xyz}{bc} + \frac{q^2xyz}{ac} + \frac{p^2xyz}{ab} \biggr] = \frac{\ell_{3D} (xyz)}{\mathfrak{L}(abc)} \biggl[ a + q^2b + p^2 c \biggr]$ $~\Rightarrow~~~ 0$ $~=$ $~ \biggl[ a + q^2b + p^2 c \biggr]\, .$

Henceforth, we will refer to this algebraic relation as the "One-Two Perpendicular Constraint."

### Arctangent

 $~\lambda_2$ $~\equiv$ $~ \tan^{-1} [x^a y^b z^c] \, .$

Then,

 $~\frac{\partial \lambda_2}{\partial x_i}$ $~=$ $~ \biggl[ \frac{1}{1+\tan^2\lambda_2} \biggr] \frac{\partial }{\partial x_i} \biggl[ x^a y^b z^c \biggr] = \cos^2\lambda_2 \cdot \frac{\partial }{\partial x_i} \biggl[ x^a y^b z^c \biggr] \, ;$

that is,

 $~\frac{\partial \lambda_2}{\partial x}$ $~=$ $~ \cos^2\lambda_2 \cdot \biggl[ x^a y^b z^c \biggr] \frac{a}{x} = \cos^2\lambda_2 \cdot \biggl[ \tan\lambda_2 \biggr] \frac{a}{x} = \sin\lambda_2 \cos\lambda_2 \biggl( \frac{a}{x} \biggr) \, ,$ $~\frac{\partial \lambda_2}{\partial y}$ $~=$ $~ \sin\lambda_2 \cos\lambda_2 \biggl( \frac{b}{y} \biggr) \, ,$ $~\frac{\partial \lambda_2}{\partial z}$ $~=$ $~ \sin\lambda_2 \cos\lambda_2 \biggl( \frac{c}{z} \biggr) \ .$

Hence,

 $~h_2^{-2}$ $~=$ $~ \sin^2\lambda_2 \cos^2\lambda_2 \biggl[ \frac{a^2}{x^2} + \frac{b^2}{y^2} + \frac{c^2}{z^2} \biggr] = \frac{\sin^2\lambda_2 \cos^2\lambda_2}{(xyz)^2} \biggl[ a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2 \biggr]$ $~\Rightarrow ~~~ h_2$ $~=$ $~ \frac{1}{\sin\lambda_2 \cos\lambda_2} \biggl[ \frac{xyz}{ \mathfrak{L} (abc)} \biggr] \, .$

And the associated unit vector is,

 $~\hat{e}_2$ $~=$ $~ \frac{1}{\mathfrak{L}} \biggl[ \hat\imath \biggl(\frac{yz}{bc}\biggr) + \hat\jmath \biggl(\frac{xz}{ac}\biggr) + \hat{k} \biggl(\frac{xy}{ab}\biggr) \biggr] \, ,$

which is the same as our default situation.

## Necessary 3rd Coordinate

The unit vector associated with the 3rd coordinate is obtained from the cross product of the first two unit vectors. That is,

 $~\hat{e}_3$ $~=$ $~\hat{e}_1 \times \hat{e}_2$ $~=$ $~ \hat\imath \biggl[ e_{1y} e_{2z} - e_{1z} e_{2y} \biggr] + \hat\jmath \biggl[ e_{1z}e_{2x} - e_{1x}e_{2z} \biggr] + \hat{k} \biggl[ e_{1x}e_{2y} - e_{1y}e_{2x} \biggr]$ $~=$ $~\frac{\ell_{3D}}{\mathfrak{L}} \biggl\{ \hat\imath \biggl[ (q^2y) \biggl( \frac{xy}{ab} \biggr) - (p^2z) \biggl( \frac{xz}{ac} \biggr) \biggr] + \hat\jmath \biggl[ (p^2z) \biggl( \frac{yz}{bc} \biggr) - (x)\biggl( \frac{xy}{ab} \biggr) \biggr] + \hat{k} \biggl[ (x) \biggl( \frac{xz}{ac} \biggr) - (q^2y) \biggl( \frac{yz}{bc} \biggr) \biggr] \biggr\}$ $~=$ $~\frac{\ell_{3D}}{\mathfrak{L}(abc)} \biggl\{ \hat\imath \biggl[ (cq^2y^2) - (b p^2z^2) \biggr]x + \hat\jmath \biggl[ (ap^2z^2) - (cx^2) \biggr]y + \hat{k} \biggl[ (bx^2) - (aq^2y^2) \biggr]z \biggr\}$

## Old Examples

### T6 Coordinates

In the set that we have elsewhere referenced as T6 coordinates, we chose: a = - 1, b = q-2, c = 0. We note, first, that this set of parameter values satisfies the above-defined One-Two Perpendicular Constraint. In this case, our generalized prescription for the 2nd coordinate generates a unit vector of the form,

 $~\hat{e}_2$ $~=$ $~ \frac{1}{\mathfrak{L}_{T6} (abc)} \biggl[ \hat\imath (ayz) + \hat\jmath (bxz) + \hat{k} (cxy) \biggr] = \frac{z}{q^2 \mathfrak{L}_{T6} (abc)} \biggl[ -\hat\imath (q^2y) + \hat\jmath (x) \biggr]$ $~=$ $~ \biggl[ -\hat\imath (q^2y) + \hat\jmath (x) \biggr] \ell_q \, ,$

where,

 $~\ell_q^{-2} \equiv \biggl[ \frac{q^4\mathfrak{L}_{T6}^2(abc)^2}{z^2}\biggr]$ $~=$ $~ \frac{q^4}{z^2}\biggl[ (yz)^2 + b^2(xz)^2 \biggr] = \biggl[ x^2 + q^4y^2 \biggr] \, .$

And it implies a unit vector for the 3rd coordinate of the form,

 $~\hat{e}_3$ $~=$ $~\frac{\ell_{3D}}{\mathfrak{L}_{T6} (abc)} \biggl\{ \hat\imath \biggl[ (cq^2y^2) - (b p^2z^2) \biggr]x + \hat\jmath \biggl[ (ap^2z^2) - (cx^2) \biggr]y + \hat{k} \biggl[ (bx^2) - (aq^2y^2) \biggr]z \biggr\}$ $~=$ $~\ell_{3D}\biggl( \frac{\ell_q q^2}{z} \biggr) \biggl\{ -\hat\imath \biggl[ \frac{p^2z^2}{q^2}\biggr] x - \hat\jmath \biggl[ p^2z^2 \biggr]y + \hat{k} \biggl[ \frac{x^2}{q^2} + q^2y^2 \biggr]z \biggr\}$ $~=$ $~\ell_q \ell_{3D} \biggl\{ -\hat\imath (x p^2z ) - \hat\jmath (q^2y p^2z) + \hat{k} (x^2 + q^4 y^2) \biggr\} \, .$

### T10 Coordinates

In the set that we have elsewhere referenced as T10 coordinates, we chose: a = 1, b = q-2, c = - 2p-2. We note, first, that this set of parameter values satisfies the above-defined One-Two Perpendicular Constraint. In this case, our generalized prescription for the 2nd coordinate generates a unit vector of the form,

 $~\hat{e}_2$ $~=$ $~ \frac{1}{\mathfrak{L}_{T10} (abc)} \biggl[ \hat\imath (ayz) + \hat\jmath (bxz) + \hat{k} (cxy) \biggr] = \frac{1}{q^2 p^2 \mathfrak{L}_{T10} (abc)} \biggl[ \hat\imath (q^2y p^2z) + \hat\jmath ( x p^2 z ) - \hat{k} ( 2xq^2y) \biggr]$

where,

 $~(abc)^2\mathfrak{L}^2_{T10}$ $~\equiv$ $~ \biggl[ a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2 \biggr] = \biggl[ y^2z^2 + \frac{x^2 z^2}{q^4} + \frac{4x^2 y^2}{p^4} \biggr]$ $~\Rightarrow~~~\mathcal{D}^2 \equiv q^4p^4(abc)^2\mathfrak{L}^2_{T10}$ $~=$ $~ \biggl[ q^4y^2 p^4z^2 + x^2 p^4z^2 + 4x^2 q^4y^2 \biggr] \, .$

And it implies a unit vector for the 3rd coordinate of the form,

 $~\hat{e}_3$ $~=$ $~\frac{\ell_{3D}}{\mathcal{D}} \biggl\{ \hat\imath \biggl[ (cq^2y^2) - (b p^2z^2) \biggr]x + \hat\jmath \biggl[ (ap^2z^2) - (cx^2) \biggr]y + \hat{k} \biggl[ (bx^2) - (aq^2y^2) \biggr]z \biggr\} q^2 p^2$ $~=$ $~\frac{\ell_{3D}}{\mathcal{D}} \biggl\{ - \hat\imath \biggl[ 2q^4y^2 + p^4z^2 \biggr]x + \hat\jmath \biggl[ p^4z^2 + 2x^2 \biggr]q^2y + \hat{k} \biggl[ x^2 - q^4y^2 \biggr]p^2z \biggr\} \, .$

## Develop 3rd-Coordinate Profile

### Setup

Reflecting back on an earlier exploration, let's define the two polynomials,

 $~\mathfrak{A} \equiv \ell_{3D}^{-2}$ $~=$ $~(x^2 + q^4y^2 + p^4z^2) \, ,$ $~\mathfrak{B} \equiv [\mathfrak{L}(abc)]^2$ $~=$ $~ [a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2] \, .$

 $~\frac{\partial \mathfrak{A}}{\partial x}$ $~=$ $~2x \, ,$ $~\frac{\partial \mathfrak{A}}{\partial y}$ $~=$ $~2q^4 y \, ,$ $~\frac{\partial \mathfrak{A}}{\partial z}$ $~=$ $~2p^4 z \, ;$ $~\frac{\partial \mathfrak{B}}{\partial x}$ $~=$ $~2x(b^2z^2 + c^2y^2) \, ,$ $~\frac{\partial \mathfrak{B}}{\partial y}$ $~=$ $~2y(a^2 z^2 + c^2x^2) \, ,$ $~\frac{\partial \mathfrak{B}}{\partial z}$ $~=$ $~2z(a^2 y^2 + b^2x^2) \, .$

Then the 3rd unit vector may be written as,

 $~[\hat{e}_3]_\mathrm{needed}$ $~=$ $~\mathfrak{A}^{-1 / 2} \mathfrak{B}^{-1 / 2} \biggl\{ \hat\imath \biggl[ (cq^2y^2) - (b p^2z^2) \biggr]x + \hat\jmath \biggl[ (ap^2z^2) - (cx^2) \biggr]y + \hat{k} \biggl[ (bx^2) - (aq^2y^2) \biggr]z \biggr\} \, .$

### Guess Third Coordinate Expression

Let's see what unit vector results if we define,

 $~\lambda_3$ $~\equiv$ $~\mathfrak{A}^{1 / 2} \mathfrak{B}^{-1 / 2} \, .$

#### Partial Derivatives

 $~\frac{\partial \lambda_3}{\partial x_i}$ $~=$ $~ \biggl[ \frac{1}{2} \mathfrak{A}^{-1 / 2} \mathfrak{B}^{-1 / 2}\biggr] \frac{\partial \mathfrak{A}}{\partial x_i} - \biggl[ \frac{1}{2} \mathfrak{A}^{1 / 2} \mathfrak{B}^{- 3 / 2} \biggr] \frac{\partial \mathfrak{B}}{\partial x_i}$ $~\Rightarrow~~~ \biggl[ \frac{2\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial x_i}$ $~=$ $~ \mathfrak{B}\cdot \frac{\partial \mathfrak{A}}{\partial x_i} - \mathfrak{A}\cdot \frac{\partial \mathfrak{B}}{\partial x_i} \, .$

First, note that,

 $~ \biggl[ \frac{2\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial x}$ $~=$ $~ \mathfrak{B} \biggl[ 2x \biggr] - \mathfrak{A}\biggl[ 2x (b^2z^2 + c^2 y^2)\biggr]$ $~=$ $~2x \biggl[ a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2 - (x^2 + q^4y^2 + p^4z^2)(b^2z^2 + c^2 y^2) \biggr]$ $~=$ $~2x \biggl[ a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2 - x^2 (b^2z^2 + c^2 y^2) - q^4y^2 (b^2z^2 + c^2 y^2) - p^4z^2(b^2z^2 + c^2 y^2) \biggr]$ $~=$ $~2x \biggl\{ (yz)^2[a^2 - q^4b^2 - c^2p^4] + (xz)^2[b^2 - b^2] + (xy)^2 [c^2 - c^2] - c^2 q^4y^4 - b^2p^4z^4 \biggr\}$ $~=$ $~2x \biggl[ y^2z^2(a^2 - q^4b^2 - c^2p^4) - c^2 q^4y^4 - b^2p^4z^4 \biggr] \, ;$
 $~ \biggl[ \frac{2\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial y}$ $~=$ $~ \mathfrak{B} \biggl[ 2q^4y \biggr] - \mathfrak{A}\biggl[ 2y(a^2 z^2 + c^2x^2) \biggr]$ $~=$ $~2y \biggl[ q^4[a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2] - (a^2 z^2 + c^2x^2) (x^2 + q^4y^2 + p^4z^2) \biggr]$ $~=$ $~2y \biggl[ a^2q^4(yz)^2 + b^2q^4(xz)^2 + c^2q^4(xy)^2 - a^2 z^2 (x^2 + q^4y^2 + p^4z^2) - c^2x^2 (x^2 + q^4y^2 + p^4z^2) \biggr]$ $~=$ $~2y \biggl\{ (yz)^2 \biggl[ a^2q^4 - a^2q^4\biggr] + (xz)^2\biggl[ b^2q^4 -a^2 - c^2p^4\biggr] + (xy)^2 \biggl[ c^2q^4 - c^2q^4 \biggr] - a^2 p^4z^4 - c^2x^4 \biggr\}$ $~=$ $~2y \biggl[ x^2z^2 ( b^2q^4 -a^2 - c^2p^4 ) - a^2 p^4z^4 - c^2x^4 \biggr] \, ;$
 $~ \biggl[ \frac{2\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial z}$ $~=$ $~ \mathfrak{B} \biggl[ 2p^4z \biggr] - \mathfrak{A}\biggl[ 2z(a^2 y^2 + b^2x^2)\biggr]$ $~=$ $~2z \biggl[ p^4[a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2] - (a^2 y^2 + b^2x^2)(x^2 + q^4y^2 + p^4z^2) \biggr]$ $~=$ $~2z \biggl[ a^2p^4(yz)^2 + b^2p^4(xz)^2 + c^2p^4(xy)^2 - a^2 y^2 (x^2 + q^4y^2 + p^4z^2) - b^2x^2(x^2 + q^4y^2 + p^4z^2) \biggr]$ $~=$ $~2z \biggl\{ (yz)^2 \biggl[ a^2p^4 - a^2p^4 \biggr] + (xz)^2 \biggl[ b^2p^4 - b^2p^4 \biggr] + (xy)^2 \biggl[c^2p^4 - a^2 - b^2q^4 \biggr] - a^2 q^4y^4 - b^2x^4 \biggr\}$ $~=$ $~2z \biggl[ (xy)^2 ( c^2p^4 - a^2 - b^2q^4 ) - a^2 q^4y^4 - b^2x^4 \biggr] \, .$

After completing a few squares, this last expression may be rewritten as …

 $~ \biggl[ \frac{2\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial z}$ $~=$ $~2z \biggl[ (xy)^2 ( c^2p^4 - a^2 - b^2q^4 ) - a^2 q^4y^4 - b^2x^4 \biggr]$ $~=$ $~2z \biggl[ (xy)^2 ( c^2p^4 - a^2 - b^2q^4 ) \pm 2abq^2(xy)^2 - (aq^2y^2 \pm bx^2)^2 \biggr]$ $~=$ $~2z \biggl[ (xy)^2 ( c^2p^4 - a^2 - b^2q^4 \pm 2abq^2) - (aq^2y^2 \pm bx^2)^2 \biggr]$ $~=$ $~2z \biggl[ (xy)^2 [ c^2p^4 -(a \pm bq^2)^2 \pm 4abq^2] - (aq^2y^2 \pm bx^2)^2 \biggr] \, .$

Now, if we choose the superior sign throughout this expression, the above-derived One-Two Perpendicular Constraint can be satisfied by setting, $~(a + bq^2) = -cp^2$. The expression then becomes,

 $~ \biggl[ \frac{2\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial z}$ $~=$ $~-2z \biggl[ (aq^2y^2 + bx^2)^2 - 4abq^2 (xy)^2 \biggr]$ $~=$ $~ -2z (aq^2y^2 - bx^2)^2 \, .$

Alternatively, suppose

 $~\lambda_3$ $~\equiv$ $~\mathfrak{A}^{m / 2} \mathfrak{B}^{-n/ 2}$ $~\Rightarrow~~~ \biggl[ \frac{2\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial x_i}$ $~=$ $~ m \mathfrak{B}\cdot \frac{\partial \mathfrak{A}}{\partial x_i} - n \mathfrak{A}\cdot \frac{\partial \mathfrak{B}}{\partial x_i} \, .$

Then we have, for example,

 $~ \frac{1}{2z}\biggl[ \frac{2\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial z}$ $~=$ $~ m \mathfrak{B} \biggl[ p^4 \biggr] - n \mathfrak{A}\biggl[ (a^2 y^2 + b^2x^2)\biggr]$ $~=$ $~ mp^4 [a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2] - n (x^2 + q^4y^2 + p^4z^2)(a^2 y^2 + b^2x^2)$ $~=$ $~ mp^4 [a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2] - n [(x^2 + q^4y^2 + p^4z^2)a^2 y^2 + (x^2 + q^4y^2 + p^4z^2)b^2x^2]$ $~=$ $~ (m-n)a^2p^4(yz)^2 + (m-n)b^2p^4(xz)^2 + (xy)^2[mc^2p^4 - na^2 - nb^2q^4] - n a^2 q^4y^4 - nb^2x^4$ $~=$ $~ (m-n)p^4[ a^2(yz)^2 + b^2(xz)^2 ] + (xy)^2[mc^2p^4 - n(a^2 + b^2q^4)] - n [ (aq^2y^2 \pm bx^2)^2 \mp 2abq^2(xy)^2] \, .$

Choosing the inferior sign then enforcing the above-derived One-Two Perpendicular Constraint by setting, $~(a + bq^2) = -cp^2$, gives,

 $~ \frac{1}{2z}\biggl[ \frac{2\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial z}$ $~=$ $~ (m-n)p^4[ a^2(yz)^2 + b^2(xz)^2 ] + (xy)^2[mc^2p^4 - n(a^2 + b^2q^4 + 2abq^2)] - n (aq^2y^2 - bx^2)^2$ $~=$ $~ (m-n)p^4[ a^2(yz)^2 + b^2(xz)^2 ] + (xy)^2[mc^2p^4 - n(-c p^2 )^2] - n (aq^2y^2 - bx^2)^2$ $~=$ $~ (m-n)p^4[ a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2 ] - n (bx^2 - aq^2y^2 )^2$ $~=$ $~ (m-n)p^4\mathfrak{B} - n (bx^2 - aq^2y^2 )^2$ $~=$ $~ m p^4\mathfrak{B} -n [p^4\mathfrak{B}+ (bx^2 - aq^2y^2 )^2 ] \, .$

Reflecting back on the first line of the "example" derivation, we recognize that,

 $~p^4\mathfrak{B}+ (bx^2 - aq^2y^2 )^2$ $~=$ $~ \mathfrak{A} (a^2 y^2 + b^2x^2)$ $~\Rightarrow ~~~(bx^2 - aq^2y^2 )^2$ $~=$ $~ \mathfrak{A} (a^2 y^2 + b^2x^2) -p^4\mathfrak{B}$

Similarly, we find,

 $~ \biggl[ \frac{2\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial x}$ $~=$ $~2x \biggl[ y^2z^2(a^2 - q^4b^2 - c^2p^4) - c^2 q^4y^4 - b^2p^4z^4 \biggr]$ $~=$ $~2x \biggl[ y^2z^2(a^2 - q^4b^2 - c^2p^4) - (cq^2y^2 \pm bp^2z^2)^2 \pm 2bcq^2y^2p^2z^2 \biggr]$ $~=$ $~2x \biggl[ (yz)^2(a^2 - q^4b^2 - c^2p^4 \pm 2bcq^2p^2) - (cq^2y^2 \pm bp^2z^2)^2 \biggr]$ $~=$ $~2x \biggl[ (yz)^2[a^2 - (bq^2 \pm cp^2)^2 \pm 4bcq^2p^2] - (cq^2y^2 \pm bp^2z^2)^2 \biggr] \, .$
 $~ \biggl[ \frac{2\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial y}$ $~=$ $~2y \biggl[ x^2z^2 ( b^2q^4 -a^2 - c^2p^4 ) - a^2 p^4z^4 - c^2x^4 \biggr]$ $~=$ $~2y \biggl[ x^2z^2 ( b^2q^4 -a^2 - c^2p^4 ) - (ap^2z^2 \pm cx^2)^2 \pm 2acx^2p^2z^2 \biggr]$ $~=$ $~2y \biggl[ x^2z^2 ( b^2q^4 -a^2 - c^2p^4 \pm 2acp^2 ) - (ap^2z^2 \pm cx^2)^2 \biggr]$ $~=$ $~2y \biggl[ (xz)^2 [ b^2q^4 -(a\pm cp^2)^2 \pm 4acp^2 ] - (ap^2z^2 \pm cx^2)^2 \biggr] \, .$

So, if we again choose the superior sign throughout these expression, the above-derived One-Two Perpendicular Constraint can be satisfied:   In the first by setting, $~(bq^2 + cp^2) = - a$; and in the second by setting, $~(a + cp^2) = - bq^2$. The expressions then become, respectively,

 $~ \biggl[ \frac{2\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial x}$ $~=$ $~ - 2x \biggl[ (cq^2y^2 + bp^2z^2)^2 - 4bcq^2 y^2 p^2 z^2 \biggr]$ $~=$ $~ - 2x (cq^2y^2 - bp^2z^2)^2 \, ;$ $~ \biggl[ \frac{2\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial y}$ $~=$ $~- 2y \biggl[ (ap^2z^2 + cx^2)^2 - 4acx^2 p^2 z^2 \biggr]$ $~=$ $~ - 2y(ap^2z^2 - cx^2)^2 \, .$
Summary

Given,

 $~\lambda_3$ $~\equiv$ $~ \mathfrak{A}^{1 / 2} \mathfrak{B}^{-1 / 2} = \biggl[ \frac{ (x^2 + q^4y^2 + p^4z^2) }{ a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2 } \biggr]^{1 / 2}$

the three relevant partial derivatives are:

 $~ \biggl[ \frac{\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial x}$ $~=$ $~ - x (cq^2y^2 - bp^2z^2)^2 \, ,$ $~ \biggl[ \frac{\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial y}$ $~=$ $~ - y(ap^2z^2 - cx^2)^2 \, ,$ $~ \biggl[ \frac{\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial z}$ $~=$ $~ -z (aq^2y^2 - bx^2)^2 \, .$

#### Scale Factor

Hence, the associated scale factor is,

 $~h_3^{-2}$ $~\equiv$ $~ \biggl( \frac{\partial \lambda_3}{\partial x} \biggr)^2 + \biggl( \frac{\partial \lambda_3}{\partial y} \biggr)^2 + \biggl( \frac{\partial \lambda_3}{\partial z} \biggr)^2$ $~$ $~=$ $~ \biggl[ \frac{\lambda_3}{\mathfrak{A}\mathfrak{B}} \biggr]^2 \biggl\{ \biggl[ - x (cq^2y^2 - bp^2z^2)^2 \biggr]^2 + \biggl[ - y(ap^2z^2 - cx^2)^2 \biggr]^2 + \biggl[ -z (aq^2y^2 - bx^2)^2 \biggr]^2 \biggr\}$ $~$ $~=$ $~ \biggl[ \frac{\lambda_3}{\mathfrak{A}\mathfrak{B}} \biggr]^2 \biggl\{ x^2 (cq^2y^2 - bp^2z^2)^4 + y^2 (ap^2z^2 - cx^2)^4 + z^2 (aq^2y^2 - bx^2)^4 \biggr\}$ $~\Rightarrow ~~~h_3$ $~=$ $~ \biggl[ \frac{\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr]\frac{1}{\mathfrak{C}} \, ,$

where,

 $~\mathfrak{C}$ $~\equiv$ $~ \biggl[ x^2 (cq^2y^2 - bp^2z^2)^4 + y^2 (ap^2z^2 - cx^2)^4 + z^2 (aq^2y^2 - bx^2)^4 \biggr]^{1 / 2} \, .$

#### Direction Cosines and Unit Vector

And the associated triplet of direction cosines is:

 $~\gamma_{31} \equiv h_3 \biggl( \frac{\partial \lambda_3}{\partial x} \biggr)$ $~=$ $~ - x (cq^2y^2 - bp^2z^2)^2\biggl[ \frac{1}{\mathfrak{C}} \biggr] \, ,$ $~\gamma_{32} \equiv h_3 \biggl( \frac{\partial \lambda_3}{\partial y} \biggr)$ $~=$ $~ - y(ap^2z^2 - cx^2)^2 \biggl[ \frac{1}{\mathfrak{C}} \biggr] \, ,$ $~ \gamma_{33} \equiv h_3 \biggl( \frac{\partial \lambda_3}{\partial y} \biggr)$ $~=$ $~ -z (aq^2y^2 - bx^2)^2 \biggl[ \frac{1}{\mathfrak{C}} \biggr] \, .$

This means that, for our particular guess of the 3rd coordinate, the relevant unit vector is,

 $~[\hat{e}_3]_\mathrm{guess}$ $~=$ $~\frac{1}{\mathfrak{C}} \biggl\{ - \hat\imath \biggl[ x (cq^2y^2 - bp^2z^2)^2 \biggr] - \hat\jmath \biggl[ y(ap^2z^2 - cx^2)^2 \biggr] - \hat{k} \biggl[ z (aq^2y^2 - bx^2)^2 \biggr] \biggr\} \, .$

### Contrast

The unit vector resulting (just derived) from our guess of the third-coordinate expression should be compared with the needed unit vector as described above, namely,

 $~[\hat{e}_3]_\mathrm{needed}$ $~=$ $~\mathfrak{A}^{-1 / 2} \mathfrak{B}^{-1 / 2} \biggl\{ \hat\imath \biggl[ x(cq^2y^2 - b p^2z^2) \biggr] + \hat\jmath \biggl[ y(ap^2z^2 - cx^2) \biggr] + \hat{k} \biggl[ z(bx^2 - aq^2y^2) \biggr] \biggr\} \, .$

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 … but no cigar!

ASIDE

Note that, $~[\hat{e}_3]_\mathrm{needed} \cdot [\hat{e}_3]_\mathrm{needed} = 1$ implies that,

 $~\mathfrak{A} \mathfrak{B} = \biggl[ \frac{(abc)\mathfrak{L}}{\ell_{3D}} \biggr]^2$ $~=$ $~ \biggl[ x(cq^2y^2 - b p^2z^2) \biggr]^2 + \biggl[ y(ap^2z^2 - cx^2) \biggr]^2 + \biggl[ z(bx^2 - aq^2y^2) \biggr]^2 \, .$

But we also know that (see, for example, immediately below),

 $~\mathfrak{A} \mathfrak{B}$ $~=$ $~ (x^2 + q^4y^2 + p^4z^2)[a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2] \, .$

What about the overall leading coefficient? That is, does $~\mathfrak{A}\mathfrak{B} = \mathfrak{C}^2$  ?  Well, given that,

 $~\mathfrak{A} = \ell_{3D}^{-2}$ $~=$ $~(x^2 + q^4y^2 + p^4z^2)$ and, $~\mathfrak{B} = (abc)^2\mathfrak{L}^2$ $~=$ $~ [a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2] \, ,$

we have,

 $~\mathfrak{A} \mathfrak{B}$ $~=$ $~ (x^2 + q^4y^2 + p^4z^2)[a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2]$ $~=$ $~ x^2 [a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2] + q^4y^2 [a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2] + p^4z^2[a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2]$ $~=$ $~x^2y^2z^2[a^2 + b^2q^4 +c^2p^4] + x^4 [b^2 z^2 + c^2 y^2] + q^4y^4 [a^2 z^2 + c^2 x^2] + p^4z^4[a^2 y^2 + b^2 x^2]$

On the other hand,

 $~\mathfrak{C}^2$ $~\equiv$ $~ x^2 (cq^2y^2 - bp^2z^2)^4 + y^2 (ap^2z^2 - cx^2)^4 + z^2 (aq^2y^2 - bx^2)^4 \, .$

### Dot With 1st Unit Vector

Is $~[\hat{e}_3]_\mathrm{guess}$ orthogonal to $~\hat{e}_1$? Let's take their dot product to see; note that, for simplicity, we will flip the sign on $~[\hat{e}_3]_\mathrm{guess}$.

 $~-\hat{e}_1 \cdot [\hat{e}_3]_\mathrm{guess} \biggl[ \frac{\mathfrak{C}}{\ell_{3D}} \biggr]$ $~=$ $~ \biggl\{ \hat\imath (x) + \hat\jmath (q^2y ) + \hat{k} (p^2 z) \biggl\} \cdot \biggl\{ \hat\imath \biggl[ x (cq^2y^2 - bp^2z^2)^2 \biggr] + \hat\jmath \biggl[ y(ap^2z^2 - cx^2)^2 \biggr] + \hat{k} \biggl[ z (aq^2y^2 - bx^2)^2 \biggr] \biggr\}$ $~=$ $~ x^2 (cq^2y^2 - bp^2z^2)^2 + q^2y^2(ap^2z^2 - cx^2)^2 + p^2z^2 (aq^2y^2 - bx^2)^2$ $~=$ $~ x^2 (c^2q^4y^4 - 2cq^2y^2 bp^2z^2 + b^2p^4z^4) + q^2y^2(a^2p^4z^4 - 2acx^2p^2z^2 + c^2x^4) + p^2z^2 (a^2q^4y^4 - 2aq^2y^2 bx^2 + b^2x^4)$ $~=$ $~ -2x^2 y^2 z^2[cq^2 bp^2 + aq^2cp^2 + aq^2 bp^2 ] + x^2 (c^2q^4y^4 + b^2p^4z^4) + q^2y^2(a^2p^4z^4 + c^2x^4) + p^2z^2 (a^2q^4y^4 + b^2x^4) \, .$

It does not appear as though the RHS of this expression can be zero for all values of the Cartesian coordinates, (x, y, z). Hence $~[\hat{e}_3]_\mathrm{guess}$ is not orthogonal to $~\hat{e}_1$.