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Point-Plane Distance

Given a Plane

ax+by+cz+d = 0
\end{displaymath} (1)

and a point $(x_0,y_0,z_0)$, the Normal to the Plane is given by
{\bf v} = \left[{\matrix{a\cr b\cr c\cr}}\right],
\end{displaymath} (2)

and a Vector from the plane to the point is given by
{\bf w} = \left[{\matrix{x-x_0\cr y-y_0\cr z-z_0\cr}}\right].
\end{displaymath} (3)

Projecting ${\bf w}$ onto ${\bf v}$,
$\displaystyle \vert{\rm proj}_{\bf v}{\bf w}\vert$ $\textstyle =$ $\displaystyle {\vert{\bf v}\cdot {\bf w}\vert\over \vert{\bf v}\vert}$  
  $\textstyle =$ $\displaystyle {\vert a(x-x_0)+b(y-y_0)+c(z-z_0)\vert\over\sqrt{a^2+b^2+c^2}}$  
  $\textstyle =$ $\displaystyle {\vert ax+by+cz-ax_0-by_0-cz_0\vert\over \sqrt{a^2+b^2+c^2}}$  
  $\textstyle =$ $\displaystyle {\vert ax_0+by_0+cz_0+d\vert\over \sqrt{a^2+b^2+c^2}}.$ (4)

© 1996-9 Eric W. Weisstein