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A plane can be taken to be a quadruple $(a, b, c, d)$. Then a point is taken to be a triple of planes $(P, Q, R)$. The point $(p, q, r)$ is valid if and only if the determinant of the matrix $A$ below is non-zero.

$$A= \left[ \begin{array}{ccc} Pa & Pb & Pc\\ Qa & Qb & Qc\\ Ra & Rb & Rc\\ \end{array} \right] $$

We can get it by simple linear algebra: The point $(P, Q, R)$ can be presented as the equation root of $Ax = b$. If this is valid then $Ax = b$ has an unique solution which implies $\det(A) \neq 0$.

Given that the point $(P, Q, R)$ is valid, it lies behind, on, or in-front of the plane $S$ if and only if the following expression is negative, zero or positve.

$$ \left\vert \begin{array}{cccc} Pa & Pb & Pc & Pd\\ Qa & Qb & Qc & Qd\\ Ra & Rb & Rc & Rd\\ Sa & Sb & Sc & Sd \end{array} \right\vert \left\vert \begin{array}{ccc} Pa & Pb & Pc\\ Qa & Qb & Qc\\ Ra & Rb & Rc \end{array} \right\vert $$

My question is: What is the mechanism behind this?

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    I suggest taking a course on linear algebra.2011-12-29
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    Can you specify what your first sentence means? Do you associate $(a, b, c, d)$ with the plane $ax + by + cz = d$? What do you mean by "valid"?2011-12-29
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    Do you associate (a,b,c,d) with the plane ax+by+cz=d? YES What do you mean by "valid"? That means (P, Q, R) is the only point shared by 3 planes.2011-12-29
  • 0
    The Question has a type error: det(A) = 0 should be det(A) != 02011-12-29
  • 0
    But for your matrix $A$ we always have $\det A=0$? Where did $d$ disappear?2011-12-29
  • 0
    Something doesn't add up. If you mean the plane $ax+by+cz=d$, then there is only a single plane. Not three planes. Also: define "in front of". Looking from where?2011-12-29
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    Jyrki, Ax=b has an unique solution which implies det(A)!=0. I've fixed the type error. Let Plane P = (Pa, Pb, Pc, Pd), Then Plane Q = (Qa, Qb, Qc, Qd), and so on.2011-12-29
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    ax + by + cz = d can also be written as (a, b, c)*(x, y, z) = d. (a, b, c) is the normal vector.2011-12-29
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    Actually we can present a 3D plane by a point p(x, y, z) and a normal vector n(nx, ny, nz). that is (px - p)*n = 0. it's equivalent to ax + by + cz = d. But have more clear geometrical significance. if a point px on the plane, dot(px - p, n) = 0. If px in front of it, dot(px - p, n) > 02011-12-29

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