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I want to fit a plane which has the following equation: $ax+by+cz+d=0$ Having a overdetermined system (more than 3 points), most people solve that equation using the normal equations (Least square). Because most people minimize over $z$-Direction, they set $c=1$ and therefore obtain the following equation $ax+by+d=-z$ which can be solved using the normal equations.

Now my question: Why can they choose $c=1$? Can I also choose $b=1$ and obtain $ax+cz+d=-y$ ? Does the selection depend on anything?

horsti
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  • Are you familiar with the term parameterization? – Andrew Chin Aug 24 '20 at 12:40
  • Not yet, could you explain what you mean in specific? – horsti Aug 24 '20 at 12:46
  • You can write a solution for $(a,b,c,d)$ where each variable is written in terms of a new variable. Or, a solution in the form $(a,b,c,d)$ can also be written as $\left(\frac ac,\frac bc,1,\frac dc\right)$. – Andrew Chin Aug 24 '20 at 12:51
  • Hmm, could you maybe explain it more theoretically why this works maybe using a system of equations..? Or do you maybe have a link, which describes that more in detail? When i look for plane parametrization google shows something in a different context i think – horsti Aug 24 '20 at 13:22
  • @horsti He divided all the terms of the equation by one of the coefficients. – Radial Arm Saw Aug 24 '20 at 13:56
  • Why is $c$ choosable when the solution vector is $(\frac{a}{c},\frac{b}{c},\frac{c}{c},\frac{d}{c})$ and not choosable when the vector is $(a,b,c,d)$ – horsti Aug 26 '20 at 08:31

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