Appendix/Ramblings/T9Coordinates

T9 Coordinates

Establish 1^st and 3^rd Coordinates

${\lambda }_1$	$\equiv$	$(x^2+q^2{y}^2+p^2{z}^2{)}^{1/2},$
${\lambda }_3$	$\equiv$	$[\frac{q^{2}y}{x^{q^2}}{]}^{1/(q^2-1)}=q^{2/(q^2-1)}{y}^{1/(q^2-1)}{x}^{q^2/(1-q^2)}.$

$\frac{\partial {\lambda }_3}{\partial x}$	$=$	$\left(\frac{q^2}{1-q^2}\right)\frac{{\lambda }_3}{x},$
$\frac{\partial {\lambda }_3}{\partial y}$	$=$	$\left(\frac{1}{q^2-1}\right)\frac{{\lambda }_3}{y},$
$h_3^{-2}$	$=$	$\left[\right(\frac{q^2}{1-q^2})\frac{{\lambda }_3}{x}{]}^2+[\left(\frac{1}{q^2-1}\right)\frac{{\lambda }_3}{y}{]}^2=\frac{{\lambda }_3^2}{(q^2-1{)}^2}\left[\frac{x^2+q^4{y}^2}{x^2{y}^2}\right]$
$\Rightarrow h_3$	$=$	$\frac{(q^2-1)}{{\lambda }_3}\left[\frac{xy}{(x^2+q^4{y}^2{)}^{1/2}}\right].$

Hence, the three direction-cosines are,

${\gamma }_{31}=h_3\left(\frac{\partial {\lambda }_3}{\partial x}\right)$	$=$	$\frac{(q^2-1)}{{\lambda }_3}\left[\frac{xy}{(x^2+q^4{y}^2{)}^{1/2}}\right]\left(\frac{q^2}{1-q^2}\right)\frac{{\lambda }_3}{x}=-\left[\frac{q^{2}y}{(x^2+q^4{y}^2{)}^{1/2}}\right],$
${\gamma }_{32}=h_3\left(\frac{\partial {\lambda }_3}{\partial y}\right)$	$=$	$\frac{(q^2-1)}{{\lambda }_3}\left[\frac{xy}{(x^2+q^4{y}^2{)}^{1/2}}\right]\left(\frac{1}{q^2-1}\right)\frac{{\lambda }_3}{y}=\left[\frac{x}{(x^2+q^4{y}^2{)}^{1/2}}\right].$

And the position vector is given by the expression,

$\overrightarrow{\mathbf{x}}$	$=$	${\hat{\mathbf{e}}}_1({\gamma }_{11}x+{\gamma }_{12}y+{\gamma }_{13}z)+{\hat{\mathbf{e}}}_2({\gamma }_{21}x+{\gamma }_{22}y+{\gamma }_{23}z)+{\hat{\mathbf{e}}}_3({\gamma }_{31}x+{\gamma }_{32}y+{\gamma }_{33}z)$
	$=$	${\hat{\mathbf{e}}}_1[x^2{\ell }_{3D}+q^2{y}^2{\ell }_{3D}+p^2{z}^2{\ell }_{3D}]+{\hat{\mathbf{e}}}_2({\gamma }_{21}x+{\gamma }_{22}y+{\gamma }_{23}z)+{\hat{\mathbf{e}}}_3[-\frac{q^{2}xy}{(x^2+q^4{y}^2{)}^{1/2}}+\frac{xy}{(x^2+q^4{y}^2{)}^{1/2}}]$
	$=$	${\hat{\mathbf{e}}}_1\left[{\lambda }_1^2{\ell }_{3D}\right]+{\hat{\mathbf{e}}}_2({\gamma }_{21}x+{\gamma }_{22}y+{\gamma }_{23}z)-{\hat{\mathbf{e}}}_3\left[\frac{(q^2-1)xy}{(x^2+q^4{y}^2{)}^{1/2}}\right].$

Guess 2^nd Coordinate

Unspecified Coefficients

${\lambda }_2$	$\equiv$	$(a{x}^2+b{y}^2+e{z}^2{)}^{-1/2}(a{x}^2+b{y}^2{)}^{1/2}$
	$=$	$[\frac{(a{x}^2+b{y}^2)}{(a{x}^2+b{y}^2+e{z}^2)}{]}^{1/2}$
	$=$	$[1+\frac{e{z}^2}{(a{x}^2+b{y}^2)}{]}^{-1/2}.$

$\frac{\partial {\lambda }_2}{\partial x}$	$=$	$-\frac{1}{2}[1+\frac{e{z}^2}{(a{x}^2+b{y}^2)}{]}^{-3/2}\frac{\partial }{\partial x}[e{z}^2(a{x}^2+b{y}^2{)}^{-1}]$
	$=$	$-\frac{e{z}^2}{2}[1+\frac{e{z}^2}{(a{x}^2+b{y}^2)}{]}^{-3/2}(-1\left)\right[(a{x}^2+b{y}^2{)}^{-2}]2ax$
	$=$	$\frac{axe{z}^2}{(a{x}^2+b{y}^2{)}^2}[1+\frac{e{z}^2}{(a{x}^2+b{y}^2)}{]}^{-3/2}$
	$=$	$\frac{axe{z}^2{\lambda }_2^3}{(a{x}^2+b{y}^2{)}^2},$
$\frac{\partial {\lambda }_2}{\partial y}$	$=$	$-\frac{1}{2}[1+\frac{e{z}^2}{(a{x}^2+b{y}^2)}{]}^{-3/2}\frac{\partial }{\partial y}[e{z}^2(a{x}^2+b{y}^2{)}^{-1}]$
	$=$	$-\frac{e{z}^2}{2}[1+\frac{e{z}^2}{(a{x}^2+b{y}^2)}{]}^{-3/2}(-1\left)\right[(a{x}^2+b{y}^2{)}^{-2}]2by$
	$=$	$\frac{bye{z}^2}{(a{x}^2+b{y}^2{)}^2}[1+\frac{e{z}^2}{(a{x}^2+b{y}^2)}{]}^{-3/2}$
	$=$	$\frac{bye{z}^2{\lambda }_2^3}{(a{x}^2+b{y}^2{)}^2},$
$\frac{\partial {\lambda }_2}{\partial z}$	$=$	$-\frac{1}{2}[1+\frac{e{z}^2}{(a{x}^2+b{y}^2)}{]}^{-3/2}\frac{\partial }{\partial z}[e{z}^2(a{x}^2+b{y}^2{)}^{-1}]$
	$=$	$-\frac{ez{\lambda }_2^3}{(a{x}^2+b{y}^2)}.$

Hence,

$h_2^{-2}$	$=$	$[\frac{axe{z}^2{\lambda }_2^3}{(a{x}^2+b{y}^2{)}^2}{]}^2+[\frac{bye{z}^2{\lambda }_2^3}{(a{x}^2+b{y}^2{)}^2}{]}^2+[\frac{ez{\lambda }_2^3}{(a{x}^2+b{y}^2)}{]}^2$
	$=$	$\left[\frac{ez{\lambda }_2^3}{(a{x}^2+b{y}^2{)}^2}{]}^2\right[(axz{)}^2+(byz{)}^2+(a{x}^2+b{y}^2{)}^2]$
$\Rightarrow h_2$	$=$	$\left[\frac{(a{x}^2+b{y}^2{)}^2}{ez{\lambda }_2^3}\right][(axz{)}^2+(byz{)}^2+(a{x}^2+b{y}^2{)}^2{]}^{-1/2}.$

The three direction-cosines are, then,

${\gamma }_{21}=h_2\left(\frac{\partial {\lambda }_2}{\partial x}\right)$	$=$	$\frac{axe{z}^2{\lambda }_2^3}{(a{x}^2+b{y}^2{)}^2}\left[\frac{(a{x}^2+b{y}^2{)}^2}{ez{\lambda }_2^3}\right][(axz{)}^2+(byz{)}^2+(a{x}^2+b{y}^2{)}^2{]}^{-1/2}$
	$=$	$axz[(axz{)}^2+(byz{)}^2+(a{x}^2+b{y}^2{)}^2{]}^{-1/2}$
${\gamma }_{22}=h_2\left(\frac{\partial {\lambda }_2}{\partial y}\right)$	$=$	$\frac{bye{z}^2{\lambda }_2^3}{(a{x}^2+b{y}^2{)}^2}\left[\frac{(a{x}^2+b{y}^2{)}^2}{ez{\lambda }_2^3}\right][(axz{)}^2+(byz{)}^2+(a{x}^2+b{y}^2{)}^2{]}^{-1/2}$
	$=$	$byz[(axz{)}^2+(byz{)}^2+(a{x}^2+b{y}^2{)}^2{]}^{-1/2}$
${\gamma }_{23}=h_2\left(\frac{\partial {\lambda }_2}{\partial z}\right)$	$=$	$-\frac{ez{\lambda }_2^3}{(a{x}^2+b{y}^2)}\left[\frac{(a{x}^2+b{y}^2{)}^2}{ez{\lambda }_2^3}\right][(axz{)}^2+(byz{)}^2+(a{x}^2+b{y}^2{)}^2{]}^{-1/2}$
	$=$	$-(a{x}^2+b{y}^2)[(axz{)}^2+(byz{)}^2+(a{x}^2+b{y}^2{)}^2{]}^{-1/2}.$

Direction Cosine Components for T9 Coordinates
$n$	${\lambda }_n$	$h_n$	$\frac{\partial {\lambda }_n}{\partial x}$	$\frac{\partial {\lambda }_n}{\partial y}$	$\frac{\partial {\lambda }_n}{\partial z}$	${\gamma }_{n1}$	${\gamma }_{n2}$	${\gamma }_{n3}$
(1)	(2)	(3)	(4)	(5)	(6)	(7)	(8)	(9)
$1$	$(x^2+q^2{y}^2+p^2{z}^2{)}^{1/2}$	${\lambda }_1{\ell }_{3D}$	$\frac{x}{{\lambda }_1}$	$\frac{q^{2}y}{{\lambda }_1}$	$\frac{p^{2}z}{{\lambda }_1}$	$(x){\ell }_{3D}$	$(q^{2}y){\ell }_{3D}$	$(p^{2}z){\ell }_{3D}$
$2A$	---	${\ell }_{3D}(x^2+q^4{y}^2{)}^{1/2}$	$\frac{x{p}^{2}z}{(x^2+q^4{y}^2)}$	$\frac{q^{2}y{p}^{2}z}{(x^2+q^4{y}^2)}$	$-1$	$\frac{x{p}^{2}z{\ell }_{3D}}{(x^2+q^4{y}^2{)}^{1/2}}$	$\frac{q^{2}y{p}^{2}z{\ell }_{3D}}{(x^2+q^4{y}^2{)}^{1/2}}$	$-\frac{(x^2+q^4{y}^2){\ell }_{3D}}{(x^2+q^4{y}^2{)}^{1/2}}$
$2B$	$[1+\frac{e{z}^2}{(a{x}^2+b{y}^2)}{]}^{-1/2}$	$\left[\frac{(a{x}^2+b{y}^2{)}^2}{ez{\lambda }_2^3}\right]{\mathfrak{𝔍}}_{2\mathrm{B}}$	$\frac{axe{z}^2{\lambda }_2^3}{(a{x}^2+b{y}^2{)}^2}$	$\frac{bye{z}^2{\lambda }_2^3}{(a{x}^2+b{y}^2{)}^2}$	$-\frac{ez{\lambda }_2^3}{(a{x}^2+b{y}^2)}$	$axz{\mathfrak{𝔍}}_{2\mathrm{B}}$	$byz{\mathfrak{𝔍}}_{2\mathrm{B}}$	$-(a{x}^2+b{y}^2){\mathfrak{𝔍}}_{2\mathrm{B}}$
$3$	$[\frac{q^{2}y}{x^{q^2}}{]}^{1/(q^2-1)}$	$\frac{(q^2-1)xy}{{\lambda }_3(x^2+q^4{y}^2{)}^{1/2}}$	$-\left(\frac{q^2}{q^2-1}\right)\frac{{\lambda }_3}{x}$	$\left(\frac{q^2}{q^2-1}\right)\frac{{\lambda }_3}{q^{2}y}$	$0$	$-\frac{q^{2}y}{(x^2+q^4{y}^2{)}^{1/2}}$	$\frac{x}{(x^2+q^4{y}^2{)}^{1/2}}$	$0$
${\ell }_{3D}$ $\equiv$ $(x^2+q^4{y}^2+p^4{z}^2{)}^{-1/2},$ ${\mathfrak{𝔍}}_0^{-2}\equiv \frac{(x^2+q^4{y}^2)}{{\ell }_{3D}^2}$ $=$ $[(x{p}^{2}z{)}^2+(q^{2}y{p}^{2}z{)}^2+(x^2+q^4{y}^2{)}^2],$ ${\mathfrak{𝔍}}_{2\mathrm{B}}$ $\equiv$ $[(axz{)}^2+(byz{)}^2+(a{x}^2+b{y}^2{)}^2{]}^{-1/2}.$

This table titled, "Direction Cosine Components for T9 Coordinates," contains the following information:

1. The (first) row labeled, $n=1$, correctly details the scale-factor, $h_1$, and the unit vector expression, ${\hat{e}}_1=(\hat{ı}{\gamma }_{11}+\hat{ȷ}{\gamma }_{12}+\hat{k}{\gamma }_{13})$, that result from the given specification of the ${\lambda }_1$ coordinate. By design, the unit vector, ${\hat{e}}_1$, is everywhere normal to the "surface" of the ellipsoid.
2. The (fourth) row labeled, $n=3$, correctly details the scale-factor, $h_3$, and the unit vector expression, ${\hat{e}}_3=(\hat{ı}{\gamma }_{31}+\hat{ȷ}{\gamma }_{32}+\hat{k}{\gamma }_{33})$, that result from the given specification of the ${\lambda }_2$ coordinate. By design, this unit vector, ${\hat{e}}_3$, has no vertical component — that is, ${\gamma }_{33}=0$ — and, by design, it is everywhere perpendicular to the "surface-normal" unit vector, ${\hat{e}}_1$.
3. We desire a unit vector, ${\hat{e}}_2$, that is mutually orthogonal to the other two unit vectors; this has been accomplished by examining their cross-product, namely, we have set ${\hat{e}}_{2\mathrm{A}}={\hat{e}}_3\times {\hat{e}}_1$. Determined in this manner, the expressions for the three direction-cosine components of ${\hat{e}}_{2\mathrm{A}}$ have been written in the last three columns of the (second) row labeled, $n=2\mathrm{A}$. While we are confident that the correct specification of ${\hat{e}}_2$ is,

   ${\hat{e}}_{2\mathrm{A}}=(\hat{ı}{\gamma }_{21}+\hat{ȷ}{\gamma }_{22}+\hat{k}{\gamma }_{23})$

   $=$

   $\hat{ı}(x{p}^{2}z){\mathfrak{𝔍}}_0+\hat{ȷ}(q^{2}y{p}^{2}z){\mathfrak{𝔍}}_0-\hat{k}(x^2+q^4{y}^2){\mathfrak{𝔍}}_0,$

   as yet (18 February 2021), we have been unable to determine an expression for the coordinate, ${\lambda }_{2\mathrm{A}}(x,y,z)$, from which all three of these direction-cosine expressions can be simultaneously derived — hence, the dashes in the second column of row 2A. The expression for $h_{2\mathrm{A}}$ that has been presented in the third column of row 2A (and framed in pink) is a guess which, when divided into each of the three direction cosines, gives respectively the three guessed partial-derivative expressions shown in columns 4, 5, and 6 of row 2A.

4. The second column of row 2B contains a guess (framed in yellow) for the coordinate expression, ${\lambda }_{2\mathrm{B}}$; this expression contains three unspecified scalar coefficients, $a$, $b$ and $e$. The remaining columns of this row contain the three partial derivatives, the associated scale factor, and the three direction cosines that result from this guessed coordinate expression. If we can find values of the three scalar coefficients that give (row 2B) expressions for the three direction cosines that perfectly match the direction cosines written in row 2A, then we will be able to state that ${\lambda }_{2\mathrm{B}}$ is — at least one form of — our sought-after third coordinate expression.

Yellow-Framed Guess 2B

Referencing the 2B table row, above, we are looking for coefficient values that map,

$axz\to x{p}^{2}z,$

$byz\to q^{2}y{p}^{2}z,$

$(a{x}^2+b{y}^2)\to (x^2+q^4{y}^2),$

and that map,

${\mathfrak{𝔍}}_{2\mathrm{B}}\equiv [(axz{)}^2+(byz{)}^2+(a{x}^2+b{y}^2{)}^2{]}^{-1/2}$

$=$

$[z^2(a^2{x}^2+b^2{y}^2)+(a{x}^2+b{y}^2{)}^2{]}^{-1/2},$

into the expression,

${\mathfrak{𝔍}}_0\equiv \frac{{\ell }_{3D}}{(x^2+q^4{y}^2{)}^{1/2}}$	$=$	$[(x^2+q^4{y}^2+p^4{z}^2)(x^2+q^4{y}^2){]}^{-1/2}$
	$=$	$[p^4{z}^2(x^2+q^4{y}^2)+(x^2+q^4{y}^2{)}^2{]}^{-1/2}$
	$=$	$[x^2{p}^4{z}^2+q^4{y}^2{p}^4{z}^2+(x^2+q^4{y}^2{)}^2{]}^{-1/2}.$

A portion of these mappings are accomplished by setting $a=p^2$ and $b=q^2{p}^2$, but this pair of specified coefficient values does not satisfy other mappings. Alternatively, a separate subset of mappings — but, again, not all mappings — is satisfied by setting $a=1$ and $b=q^4$. So the yellow-framed guess 2B does not provide a correct second-coordinate expression.

Blue-Framed Guess 2C

Let's try,

${\lambda }_{2\mathrm{B}}$

$=$

$(cz{)}^{2e}(a{x}^2+b{y}^2{)}^{-f}.$

Direction Cosine Components for Additional T9 Coordinate Guesses
$n$   (1)	${\lambda }_n$   (2)	$h_n$   (3)	$\partial {\lambda }_n/\partial x$   (4)	$\partial {\lambda }_n/\partial y$   (5)	$\partial {\lambda }_n/\partial z$   (6)	${\gamma }_{n1}$   (7)	${\gamma }_{n2}$   (8)	${\gamma }_{n3}$   (9)
$2A$	---	---	---	---	---	$x{p}^{2}z{\mathfrak{𝔍}}_0$	$q^{2}y{p}^{2}z{\mathfrak{𝔍}}_0$	$-(x^2+q^4{y}^2){\mathfrak{𝔍}}_0$
$2B$	$[1+\frac{e{z}^2}{(a{x}^2+b{y}^2)}{]}^{-1/2}$	$\left[\frac{(a{x}^2+b{y}^2{)}^2}{ez{\lambda }_2^3}\right]{\mathfrak{𝔍}}_{2\mathrm{B}}$	$\frac{axe{z}^2{\lambda }_2^3}{(a{x}^2+b{y}^2{)}^2}$	$\frac{bye{z}^2{\lambda }_2^3}{(a{x}^2+b{y}^2{)}^2}$	$-\frac{ez{\lambda }_2^3}{(a{x}^2+b{y}^2)}$	$axz{\mathfrak{𝔍}}_{2\mathrm{B}}$	$byz{\mathfrak{𝔍}}_{2\mathrm{B}}$	$-(a{x}^2+b{y}^2){\mathfrak{𝔍}}_{2\mathrm{B}}$
$2C$	$(cz{)}^{2e}(a{x}^2+b{y}^2{)}^{-f}$	$\left[\frac{z(a{x}^2+b{y}^2)}{2e{\lambda }_{2\mathrm{C}}}\right]{\mathfrak{𝔍}}_{2\mathrm{C}}$	$-\frac{2afx{\lambda }_{2\mathrm{C}}}{(a{x}^2+b{y}^2)}$	$-\frac{2bfy{\lambda }_{2\mathrm{C}}}{(a{x}^2+b{y}^2)}$	$\frac{2e{\lambda }_{2\mathrm{C}}}{z}$	$-\left(\frac{afxz}{e}\right){\mathfrak{𝔍}}_{2\mathrm{C}}$	$-\left(\frac{bfyz}{e}\right){\mathfrak{𝔍}}_{2\mathrm{C}}$	$(a{x}^2+b{y}^2){\mathfrak{𝔍}}_{2\mathrm{C}}$
${\ell }_{3D}$ $\equiv$ $(x^2+q^4{y}^2+p^4{z}^2{)}^{-1/2},$ ${\mathfrak{𝔍}}_0^{-2}\equiv \frac{(x^2+q^4{y}^2)}{{\ell }_{3D}^2}$ $=$ $[(x{p}^{2}z{)}^2+(q^{2}y{p}^{2}z{)}^2+(x^2+q^4{y}^2{)}^2],$ ${\mathfrak{𝔍}}_{2\mathrm{B}}$ $\equiv$ $[(axz{)}^2+(byz{)}^2+(a{x}^2+b{y}^2{)}^2{]}^{-1/2},$ ${\mathfrak{𝔍}}_{2\mathrm{C}}$ $\equiv$ $[(xz{)}^2(\frac{fa}{e}{)}^2+(yz{)}^2(\frac{fb}{e}{)}^2+(a{x}^2+b{y}^2{)}^2{]}^{-1/2}.$

Examining just the expression for ${\mathfrak{𝔍}}_{2\mathrm{C}}$, we see that we definitely need: $a=1$ and $b=q^4$. Also, we need,

$\frac{af}{e}$

$=$

$p^2$

$\Rightarrow$

$\frac{f}{e}$

$=$

$p^2;$

and, separately we need,

$\frac{fb}{e}$

$=$

$q^2{p}^2$

$\Rightarrow$

$\frac{f}{e}$

$=$

$\frac{p^2}{q^2}.$

These cannot simultaneously be satisfied. So the blue-framed guess 2C does not provide a correct second-coordinate expression. But we are very close! We need one additional scalar coefficient degree of freedom.

Next Thought

More generally, this leads to an expression for the scale-factor of the form,

$h_2^{-2}$	$\sim$	$\mathfrak{𝔊}(x,y,z)[x^2{p}^4{z}^2+q^4{y}^2{p}^4{z}^2+(x^2+q^4{y}^2{)}^2]$
$\Rightarrow \frac{\partial {\lambda }_2}{\partial x}$	$\sim$	$[\mathfrak{𝔊}(x,y,z){]}^{1/2}[x{p}^{2}z];$		$\frac{\partial {\lambda }_2}{\partial y}$	$\sim$	$[\mathfrak{𝔊}(x,y,z){]}^{1/2}[q^{2}y{p}^{2}z];$		$\frac{\partial {\lambda }_2}{\partial z}$	$\sim$	$[\mathfrak{𝔊}(x,y,z){]}^{1/2}[x^2+q^4{y}^2].$

---

Now, if we set, $a=p^2$ and $b=q^2{p}^2$, we have,

$(axz{)}^2+(byz{)}^2+(a{x}^2+b{y}^2{)}^2$	$=$	$(x{p}^{2}z{)}^2+(q^2{p}^{2}yz{)}^2+(p^2{x}^2+q^2{p}^2{y}^2{)}^2$
	$=$	$p^4[q^4{y}^2{z}^2+x^2{z}^2+(x^2+q^2{y}^2{)}^2],$

in which case,

Complete Orthogonality Check

If the set of unit vectors is indeed orthogonal, then we must find that,

${\gamma }_{mn}$

$=$

$M_{mn},$

where the quantity $M_{mn}$ is the minor of ${\gamma }_{mn}$ in the determinant, $|{\gamma }_{mn}|$. (Note: This last expression is true only for right-handed coordinate systems. If the coordinate system is left-handed, we should find, ${\gamma }_{mn}=-M_{mn}$.) More specifically, for any right-handed, orthogonal curvilinear coordinate system we should find:

${\gamma }_{11}$ $=$ ${\gamma }_{22}{\gamma }_{33}-{\gamma }_{23}{\gamma }_{32},$ ${\gamma }_{12}$ $=$ ${\gamma }_{23}{\gamma }_{31}-{\gamma }_{21}{\gamma }_{33},$ ${\gamma }_{13}$ $=$ ${\gamma }_{21}{\gamma }_{32}-{\gamma }_{22}{\gamma }_{31},$

${\gamma }_{21}$ $=$ ${\gamma }_{32}{\gamma }_{13}-{\gamma }_{33}{\gamma }_{12},$ ${\gamma }_{22}$ $=$ ${\gamma }_{33}{\gamma }_{11}-{\gamma }_{31}{\gamma }_{13},$ ${\gamma }_{23}$ $=$ ${\gamma }_{31}{\gamma }_{12}-{\gamma }_{32}{\gamma }_{11},$

${\gamma }_{31}$ $=$ ${\gamma }_{12}{\gamma }_{23}-{\gamma }_{13}{\gamma }_{22},$ ${\gamma }_{32}$ $=$ ${\gamma }_{13}{\gamma }_{21}-{\gamma }_{11}{\gamma }_{23},$ ${\gamma }_{33}$ $=$ ${\gamma }_{11}{\gamma }_{22}-{\gamma }_{12}{\gamma }_{21}.$

and the position vector is,

$\overrightarrow{\mathbf{x}}$	$=$	$\hat{\mathit{ı}}x+\hat{\mathit{ȷ}}y+\hat{\mathbf{k}}z$
	$=$	${\hat{\mathbf{e}}}_1({\gamma }_{11}x+{\gamma }_{12}y+{\gamma }_{13}z)+{\hat{\mathbf{e}}}_2({\gamma }_{21}x+{\gamma }_{22}y+{\gamma }_{23}z)+{\hat{\mathbf{e}}}_3({\gamma }_{31}x+{\gamma }_{32}y+{\gamma }_{33}z).$

For (κ₁, κ₂, κ₃)

${\gamma }_{22}{\gamma }_{33}-{\gamma }_{23}{\gamma }_{32}$	$=$	$\left[\frac{q^{2}y{p}^{2}z{\ell }_{3D}}{(x^2+q^4{y}^2{)}^{1/2}}\right]\left[0\right]+\left[\frac{(x^2+q^4{y}^2){\ell }_{3D}}{(x^2+q^4{y}^2{)}^{1/2}}\right]\left[\frac{x}{(x^2+q^4{y}^2{)}^{1/2}}\right]=x{\ell }_{3D}={\gamma }_{11}.$       Yes!
${\gamma }_{23}{\gamma }_{31}-{\gamma }_{21}{\gamma }_{33}$	$=$	$[(x^2+q^4{y}^2{)}^{1/2}{\ell }_{3D}\left]\right[\frac{q^{2}y}{(x^2+q^4{y}^2{)}^{1/2}}]-[\frac{x{p}^{2}z{\ell }_{3D}}{(x^2+q^4{y}^2{)}^{1/2}}\left]\right[0]=q^{2}y{\ell }_{3D}={\gamma }_{12}.$       Yes!
${\gamma }_{21}{\gamma }_{32}-{\gamma }_{22}{\gamma }_{31}$	$=$	$\left[\frac{x{p}^{2}z{\ell }_{3D}}{(x^2+q^4{y}^2{)}^{1/2}}\right]\left[\frac{x}{(x^2+q^4{y}^2{)}^{1/2}}\right]+\left[\frac{q^{2}y{p}^{2}z{\ell }_{3D}}{(x^2+q^4{y}^2{)}^{1/2}}\right]\left[\frac{q^{2}y}{(x^2+q^4{y}^2{)}^{1/2}}\right]$
	$=$	$\left[\frac{x^2{p}^{2}z{\ell }_{3D}}{(x^2+q^4{y}^2)}\right]+\left[\frac{q^4{y}^2{p}^{2}z{\ell }_{3D}}{(x^2+q^4{y}^2)}\right]=p^{2}z{\ell }_{3D}={\gamma }_{13}.$       Yes!

${\gamma }_{32}{\gamma }_{13}-{\gamma }_{33}{\gamma }_{12}$	$=$	$\left[\frac{x}{(x^2+q^4{y}^2{)}^{1/2}}\right]\left[p^{2}z{\ell }_{3D}\right]-\left[0\right]\left[q^{2}y{\ell }_{3D}\right]=\left[\frac{x{p}^{2}z{\ell }_{3D}}{(x^2+q^4{y}^2{)}^{1/2}}\right]={\gamma }_{21}.$       Yes!
${\gamma }_{33}{\gamma }_{11}-{\gamma }_{31}{\gamma }_{13}$	$=$	$\left[0\right]\left[x{\ell }_{3D}\right]+\left[\frac{q^{2}y}{(x^2+q^4{y}^2{)}^{1/2}}\right]\left[p^{2}z{\ell }_{3D}\right]=\left[\frac{q^{2}y{p}^{2}z{\ell }_{3D}}{(x^2+q^4{y}^2{)}^{1/2}}\right]={\gamma }_{22}.$       Yes!
${\gamma }_{31}{\gamma }_{12}-{\gamma }_{32}{\gamma }_{11}$	$=$	$-\left[\frac{q^{2}y}{(x^2+q^4{y}^2{)}^{1/2}}\right]\left[q^{2}y{\ell }_{3D}\right]-\left[\frac{x}{(x^2+q^4{y}^2{)}^{1/2}}\right]\left[x{\ell }_{3D}\right]=-(x^2+q^4{y}^2{)}^{1/2}{\ell }_{3D}={\gamma }_{23}.$       Yes!

${\gamma }_{12}{\gamma }_{23}-{\gamma }_{13}{\gamma }_{22}$	$=$	$-\left[q^{2}y{\ell }_{3D}\right][(x^2+q^4{y}^2{)}^{1/2}{\ell }_{3D}]-[p^{2}z{\ell }_{3D}\left]\right[\frac{q^{2}y{p}^{2}z{\ell }_{3D}}{(x^2+q^4{y}^2{)}^{1/2}}]$
	$=$	$-\frac{q^{2}y{\ell }_{3D}^2}{(x^2+q^4{y}^2{)}^{1/2}}[x^2+q^4{y}^2+p^4{z}^2]=-\frac{q^{2}y}{(x^2+q^4{y}^2{)}^{1/2}}={\gamma }_{31}.$       Yes!
${\gamma }_{13}{\gamma }_{21}-{\gamma }_{11}{\gamma }_{23}$	$=$	$\left[p^{2}z{\ell }_{3D}\right]\left[\frac{x{p}^{2}z{\ell }_{3D}}{(x^2+q^4{y}^2{)}^{1/2}}\right]+\left[x{\ell }_{3D}\right][(x^2+q^4{y}^2{)}^{1/2}{\ell }_{3D}]$
	$=$	$\frac{x{\ell }_{3D}^2}{(x^2+q^4{y}^2{)}^{1/2}}[x^2+q^4{y}^2+p^4{z}^2]=\frac{x}{(x^2+q^4{y}^2{)}^{1/2}}={\gamma }_{32}.$       Yes!
${\gamma }_{11}{\gamma }_{22}-{\gamma }_{12}{\gamma }_{21}$	$=$	$\left[x{\ell }_{3D}\right]\left[\frac{q^{2}y{p}^{2}z{\ell }_{3D}}{(x^2+q^4{y}^2{)}^{1/2}}\right]-\left[q^{2}y{\ell }_{3D}\right]\left[\frac{x{p}^{2}z{\ell }_{3D}}{(x^2+q^4{y}^2{)}^{1/2}}\right]=0={\gamma }_{33}.$       Yes!

Given that the prescribed interrelationships between all nine direction cosines are satisfied, we conclude that the $({\kappa }_1,{\kappa }_2,{\kappa }_3)$ coordinate system is an orthogonal one. Accordingly, the position vector is,

$\overrightarrow{\mathbf{x}}$	$=$	${\hat{\mathbf{e}}}_1\{x^2{\ell }_{3D}+q^2{y}^2{\ell }_{3D}+p^2{z}^2{\ell }_{3D}\}$
		$+{\hat{\mathbf{e}}}_2\{x^2+q^2{y}^2-\frac{1}{p^2}(x^2+q^4{y}^2)\}\frac{p^{2}z{\ell }_{3D}}{(x^2+q^4{y}^2{)}^{1/2}}$
		$+{\hat{\mathbf{e}}}_3\{-\frac{x{q}^{2}y}{(x^2+q^4{y}^2{)}^{1/2}}+\frac{xy}{(x^2+q^4{y}^2{)}^{1/2}}\}$
	$=$	${\hat{\mathbf{e}}}_1\left[{\kappa }_1^2{\ell }_{3D}\right]+{\hat{\mathbf{e}}}_2[x^2+q^2{y}^2-\frac{1}{p^2}(x^2+q^4{y}^2)]\frac{p^{2}z{\ell }_{3D}}{(x^2+q^4{y}^2{)}^{1/2}}+{\hat{\mathbf{e}}}_3\left[\frac{(1-q^2)xy}{(x^2+q^4{y}^2{)}^{1/2}}\right]$
	$=$	${\hat{\mathbf{e}}}_1\left[{\kappa }_1^2{\ell }_{3D}\right]+{\hat{\mathbf{e}}}_2\left[x^2\right(1-\frac{1}{p^2})-q^2{y}^2(q^2-1\left)\right]\frac{p^{2}z{\ell }_{3D}}{(x^2+q^4{y}^2{)}^{1/2}}-{\hat{\mathbf{e}}}_3\left[\frac{(q^2-1)xy}{(x^2+q^4{y}^2{)}^{1/2}}\right].$

For (κ₁, κ₄, κ₅)

${\gamma }_{42}{\gamma }_{53}-{\gamma }_{43}{\gamma }_{52}$	$=$	$\left[\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}\right(\frac{1}{q^{2}y}\left)\right]\left[\frac{p^{2}z{\ell }_{3D}}{{𝒟}}\right(x^2-q^4{y}^2\left)\right]+\left[\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}\right(\frac{2}{p^{2}z}\left)\right]\left[\frac{q^{2}y{\ell }_{3D}}{{𝒟}}\right(p^4{z}^2+2x^2\left)\right]$
	$=$	$\frac{x{\ell }_{3D}}{{{𝒟}}^2}[(x^2-q^4{y}^2)p^4{z}^2+(p^4{z}^2+2x^2)2q^4{y}^2]$
	$=$	$\frac{x{\ell }_{3D}}{{{𝒟}}^2}[x^2{p}^4{z}^2+4x^2{q}^4{y}^2+q^4{y}^2{p}^4{z}^2]=x{\ell }_{3D}={\gamma }_{11}.$       Yes!
${\gamma }_{43}{\gamma }_{51}-{\gamma }_{41}{\gamma }_{53}$	$=$	$\left[\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}\right(\frac{2}{p^{2}z}\left)\right]\left[\frac{x{\ell }_{3D}}{{𝒟}}\right(2q^4{y}^2+p^4{z}^2\left)\right]-\left[\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}\right(\frac{1}{x}\left)\right]\left[\frac{p^{2}z{\ell }_{3D}}{{𝒟}}\right(x^2-q^4{y}^2\left)\right]$
	$=$	$\frac{q^{2}y{\ell }_{3D}}{{{𝒟}}^2}[2x^2(2q^4{y}^2+p^4{z}^2)-p^4{z}^2(x^2-q^4{y}^2)]$
	$=$	$\frac{q^{2}y{\ell }_{3D}}{{{𝒟}}^2}[4x^2{q}^4{y}^2+x^2{p}^4{z}^2+q^4{y}^2{p}^4{z}^2]=q^{2}y{\ell }_{3D}={\gamma }_{12}.$       Yes!
${\gamma }_{41}{\gamma }_{52}-{\gamma }_{42}{\gamma }_{51}$	$=$	$\left[\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}\right(\frac{1}{x}\left)\right]\left[\frac{q^{2}y{\ell }_{3D}}{{𝒟}}\right(p^4{z}^2+2x^2\left)\right]+\left[\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}\right(\frac{1}{q^{2}y}\left)\right]\left[\frac{x{\ell }_{3D}}{{𝒟}}\right(2q^4{y}^2+p^4{z}^2\left)\right]$
	$=$	$\frac{p^{2}z{\ell }_{3D}}{{{𝒟}}^2}[q^4{y}^2(p^4{z}^2+2x^2)+x^2(2q^4{y}^2+p^4{z}^2)]$
	$=$	$\frac{p^{2}z{\ell }_{3D}}{{{𝒟}}^2}[q^4{y}^2{p}^4{z}^2+4x^2{q}^4{y}^2+x^2{p}^4{z}^2]=p^{2}z{\ell }_{3D}={\gamma }_{13}.$       Yes!

${\gamma }_{52}{\gamma }_{13}-{\gamma }_{53}{\gamma }_{12}$	$=$	$\left[\frac{q^{2}y{p}^{2}z{\ell }_{3D}^2}{{𝒟}}\right(p^4{z}^2+2x^2\left)\right]-\left[\frac{q^{2}y{p}^{2}z{\ell }_{3D}^2}{{𝒟}}\right(x^2-q^4{y}^2\left)\right]$
	$=$	$\frac{q^{2}y{p}^{2}z{\ell }_{3D}^2}{{𝒟}}[x^2+q^4{y}^2+p^4{z}^2]=\frac{q^{2}y{p}^{2}z}{{𝒟}}={\gamma }_{41}.$       Yes!
${\gamma }_{53}{\gamma }_{11}-{\gamma }_{51}{\gamma }_{13}$	$=$	$\left[\frac{x{p}^{2}z{\ell }_{3D}^2}{{𝒟}}\right(x^2-q^4{y}^2\left)\right]+\left[\frac{x{p}^{2}z{\ell }_{3D}^2}{{𝒟}}\right(2q^4{y}^2+p^4{z}^2\left)\right]$
	$=$	$\frac{x{p}^{2}z{\ell }_{3D}^2}{{𝒟}}[x^2+q^4{y}^2+p^4{z}^2]=\frac{x{p}^{2}z}{{𝒟}}={\gamma }_{42}.$       Yes!
${\gamma }_{51}{\gamma }_{12}-{\gamma }_{52}{\gamma }_{11}$	$=$	$-\left[\frac{x{q}^{2}y{\ell }_{3D}^2}{{𝒟}}\right(2q^4{y}^2+p^4{z}^2\left)\right]-\left[\frac{x{q}^{2}y{\ell }_{3D}^2}{{𝒟}}\right(p^4{z}^2+2x^2\left)\right]$
	$=$	$-\frac{2x{q}^{2}y{\ell }_{3D}^2}{{𝒟}}[x^2+q^4{y}^2+p^4{z}^2]=-\frac{2x{q}^{2}y}{{𝒟}}={\gamma }_{43}.$       Yes!

${\gamma }_{12}{\gamma }_{43}-{\gamma }_{13}{\gamma }_{42}$	$=$	$-\left[\frac{2x{q}^4{y}^2{\ell }_{3D}}{{𝒟}}\right]-\left[\frac{x{p}^4{z}^2{\ell }_{3D}}{{𝒟}}\right]=-\left[\frac{x{\ell }_{3D}}{{𝒟}}\right](2q^4{y}^2+p^4{z}^2)={\gamma }_{51}.$       Yes!
${\gamma }_{13}{\gamma }_{41}-{\gamma }_{11}{\gamma }_{43}$	$=$	$\left[\frac{q^{2}y{p}^4{z}^2{\ell }_{3D}}{{𝒟}}\right]+\left[\frac{2x^2{q}^{2}y{\ell }_{3D}}{{𝒟}}\right]=\left[\frac{q^{2}y{\ell }_{3D}}{{𝒟}}\right](p^4{z}^2+2x^2)={\gamma }_{52}.$       Yes!
${\gamma }_{11}{\gamma }_{42}-{\gamma }_{12}{\gamma }_{41}$	$=$	$\left[\frac{x^2{p}^{2}z{\ell }_{3D}}{{𝒟}}\right]-\left[\frac{q^4{y}^2{p}^{2}z{\ell }_{3D}}{{𝒟}}\right]=\left[\frac{p^{2}z{\ell }_{3D}}{{𝒟}}\right](x^2-q^4{y}^2)={\gamma }_{53}.$       Yes!

Given that the prescribed interrelationships between all nine direction cosines are satisfied, we conclude that the $({\kappa }_1,{\kappa }_4,{\kappa }_5)$ coordinate system is an orthogonal one. Accordingly, the position vector is,

$\overrightarrow{\mathbf{x}}$	$=$	${\hat{\mathbf{e}}}_1({\gamma }_{11}x+{\gamma }_{12}y+{\gamma }_{13}z)+{\hat{\mathbf{e}}}_2({\gamma }_{41}x+{\gamma }_{42}y+{\gamma }_{43}z)+{\hat{\mathbf{e}}}_3({\gamma }_{51}x+{\gamma }_{52}y+{\gamma }_{53}z)$
	$=$	${\hat{\mathbf{e}}}_1\{x^2{\ell }_{3D}+q^2{y}^2{\ell }_{3D}+p^2{z}^2{\ell }_{3D}\}$
		$+{\hat{\mathbf{e}}}_2\{\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}+\frac{xy{p}^{2}z}{{𝒟}}-\frac{2x{q}^{2}yz}{{𝒟}}\}$
		$+{\hat{\mathbf{e}}}_3\{-\frac{{\ell }_{3D}{x}^2}{{𝒟}}(2q^4{y}^2+p^4{z}^2)+\frac{{\ell }_{3D}{q}^2{y}^2}{{𝒟}}(p^4{z}^2+2x^2)+\frac{{\ell }_{3D}{p}^2{z}^2}{{𝒟}}(x^2-q^4{y}^2)\}$
	$=$	${\hat{\mathbf{e}}}_1\left\{{\kappa }_1^2{\ell }_{3D}\right\}+{\hat{\mathbf{e}}}_2\{q^2{p}^2+p^2-2q^2\}\frac{xyz}{{𝒟}}$
		$+{\hat{\mathbf{e}}}_3\{-x^2(2q^4{y}^2+p^4{z}^2)+q^2{y}^2(p^4{z}^2+2x^2)+p^2{z}^2(x^2-q^4{y}^2)\}\frac{{\ell }_{3D}}{{𝒟}}$
	$=$	${\hat{\mathbf{e}}}_1\left\{{\kappa }_1^2{\ell }_{3D}\right\}+{\hat{\mathbf{e}}}_2\{q^2{p}^2+p^2-2q^2\}\frac{xyz}{{𝒟}}-{\hat{\mathbf{e}}}_3\{2x^2{q}^2{y}^2(q^2-1)+x^2{p}^2{z}^2(p^2-1)+q^2{y}^2{p}^2{z}^2(q^2-p^2)\}\frac{{\ell }_{3D}}{{𝒟}}.$

See Also

• Direction Cosines
• Trials up through T7 Coordinates