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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 in vector form ( is a unit vector) and a point p, the distance from a point to a line from p to l is .

In this case, point P is at .

Line BC has and .

Using the perpendicular distance formula,

Substituting K = a + b + c,

.

Similar calculation on lines AC and AB gives

and .

This shows that the distance of the point from the respective lines is linearly proportional to the original values a, b and c.[1]
  1. Ternary plots
Source Own work
Author Cmglee

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Date/TimeThumbnailDimensionsUserComment
current00:03, 24 June 2013Thumbnail for version as of 00:03, 24 June 2013512 × 128 (5 KB)CmgleeFix svg viewBox.
00:01, 24 June 2013Thumbnail for version as of 00:01, 24 June 2013512 × 512 (5 KB)CmgleeFinish graphic.
19:58, 21 June 2013Thumbnail for version as of 19:58, 21 June 2013512 × 154 (6 KB)Cmglee{{Information |Description ={{en|1=Derivation of ternary plots from Cartesian coordinates: For a line '''l''' = '''s''' + ''t'' '''n''' in vector form, where '''n''' is a unit vector, and a point '''p''', the [[distance from a point to a line|perp...

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