English: Derivation of a ternary plot from Cartesian coordinates:

Figure (1) shows an oblique projection of point P(a,b,c) in a 3-dimensional Cartesian space with axes a, b and c, respectively.

If a + b + c = K (a positive constant), P is restricted to a plane containing A(K,0,0), B(0,K,0) and C(0,0,K). If a, b and c each cannot be negative, P is restricted to the triangle bounded by A, B and C, as in (2).

In (3), the axes are rotated to give an isometric view. The triangle, viewed face-on, appears equilateral.

In (4), the distances of P from lines BC, AC and AB are denoted by a' , b' and c' , respectively.

For any line l = s + t n̂ in vector form (n̂ is a unit vector) and a point p, the distance from a point to a line from p to l is ${\displaystyle \|(\mathbf {s} -\mathbf {p} )-((\mathbf {s} -\mathbf {p} )\cdot \mathbf {\hat {n}} )\mathbf {\hat {n}} \|}$ .

In this case, point P is at ${\displaystyle \mathbf {p} ={\begin{pmatrix}a\\b\\c\end{pmatrix}}}$ .

Line BC has ${\displaystyle \mathbf {s} ={\begin{pmatrix}0\\K\\0\end{pmatrix}}}$ and ${\displaystyle \mathbf {\hat {n}} ={\frac {{\Big (}{\begin{smallmatrix}0\\K\\0\end{smallmatrix}}{\Big )}-{\Big (}{\begin{smallmatrix}0\\0\\K\end{smallmatrix}}{\Big )}}{{\Big |}{\Big |}{\Big (}{\begin{smallmatrix}0\\K\\0\end{smallmatrix}}{\Big )}-{\Big (}{\begin{smallmatrix}0\\0\\K\end{smallmatrix}}{\Big )}{\Big |}{\Big |}}}={\frac {{\Big (}{\begin{smallmatrix}0\\K\\-K\end{smallmatrix}}{\Big )}}{\sqrt {0^{2}+K^{2}+(-K)^{2}}}}={\begin{pmatrix}0\\\;\;1/{\sqrt {2}}\\-1/{\sqrt {2}}\end{pmatrix}}}$ .

Using the perpendicular distance formula,

${\displaystyle {\begin{aligned}a'&={\bigg |}{\bigg |}{\Big (}{\begin{smallmatrix}-a\\K-b\\-c\end{smallmatrix}}{\Big )}-{\bigg (}{\Big (}{\begin{smallmatrix}-a\\K-b\\-c\end{smallmatrix}}{\Big )}\cdot {\Big (}{\begin{smallmatrix}0\\\;\;1/{\sqrt {2}}\\-1/{\sqrt {2}}\end{smallmatrix}}{\Big )}{\bigg )}{\Big (}{\begin{smallmatrix}0\\\;\;1/{\sqrt {2}}\\-1/{\sqrt {2}}\end{smallmatrix}}{\Big )}{\bigg |}{\bigg |}\\&={\bigg |}{\bigg |}{\Big (}{\begin{smallmatrix}-a\\K-b\\-c\end{smallmatrix}}{\Big )}-{\Big (}0+{\tfrac {K-b}{\sqrt {2}}}+{\tfrac {c}{\sqrt {2}}}{\Big )}{\Big (}{\begin{smallmatrix}0\\\;\;1/{\sqrt {2}}\\-1/{\sqrt {2}}\end{smallmatrix}}{\Big )}{\bigg |}{\bigg |}\\&={\bigg |}{\bigg |}{\bigg (}{\begin{smallmatrix}-a\\K-b-{\tfrac {K-b+c}{2}}\\-c+{\tfrac {K-b+c}{2}}\end{smallmatrix}}{\bigg )}{\bigg |}{\bigg |}={\bigg |}{\bigg |}{\bigg (}{\begin{smallmatrix}-a\\{\tfrac {K-b-c}{2}}\\{\tfrac {K-b-c}{2}}\end{smallmatrix}}{\bigg )}{\bigg |}{\bigg |}\\&={\sqrt {(-a)^{2}+{\big (}{\tfrac {K-b-c}{2}}{\big )}^{2}+{\big (}{\tfrac {K-b-c}{2}}{\big )}^{2}}}={\sqrt {a^{2}+{\tfrac {(K-b-c)^{2}}{2}}}}\\\end{aligned}}}$

Substituting K = a + b + c,

${\displaystyle a'={\sqrt {a^{2}+{\tfrac {(a+b+c-b-c)^{2}}{2}}}}={\sqrt {a^{2}+{\tfrac {a^{2}}{2}}}}=a{\sqrt {\tfrac {3}{2}}}}$ .

Similar calculation on lines AC and AB gives

${\displaystyle b'=b{\sqrt {\tfrac {3}{2}}}}$ and ${\displaystyle c'=c{\sqrt {\tfrac {3}{2}}}}$ .

This shows that the distance of the point from the respective lines is linearly proportional to the original values a, b and c.^[1]