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I am studying the problem of the solution of the following equation: a*P+(1-a)*R=b*Q+(1-b)*T, wher P,R,Q and T are given points in the space. My unknows are a and b, so I have an overdetermined system of 3 equations (one for each coordinate) and 2 unknows (a and b). I have found that the 4 points R,Q,P and T are coplanar, and the book I have studying adfirm that this information assures that the solution of the system is unique, but I don't understand why. Thanks.

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  • $\begingroup$ We need more informations the equation is not solvable with the information you gave us, try $P=R=0$ and $Q=e_1$ and $T=e_2$, it has no solution $\endgroup$ – Dominic Michaelis Apr 11 '13 at 11:35
  • $\begingroup$ The context is that of the joint between two surfaces and I want that the surfaces meet with a G1 continuity along a common curve. So we have P,Q and T wich gave us the tangent plane of the common curve in a point of the curve, and R,Q and T give the tangent plane calculated from the other patch. We want that this plane varies with continuity along the boundary curve, so that the 4 distinct point are always coplanar. $\endgroup$ – dfretghu Apr 11 '13 at 11:50
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The reason it has a solution when the points are coplanar is this. The points may all be written as a combination of a basis for that plane such as \begin{align} P &= \mathbf{x}p_x + \mathbf{y}p_y \\ Q &= \mathbf{x}q_x + \mathbf{y}q_y \\ R &= \mathbf{x}r_x + \mathbf{y}r_y \\ T &= \mathbf{x}t_x + \mathbf{y}t_y \\ \end{align} where the basis is the vectors $\mathbf{x}$ and $\mathbf{y}$.

Once each point is in terms of those two, the equation system is written as two equations in two unknowns. The unkowns are $a$ and $b$ of course, and the equations are: \begin{align} p_x a + r_x(1-a) &= q_x b + t_x (1-b) \\ p_y a + r_y(1-a) &= q_y b + t_y (1-b) \\ \end{align}

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