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Mathematics-Online course: Prepcourse Mathematics - Linear Algebra and Geometry - Lines and Planes

Distance Point-Plane

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Given a plane with normal vector $ \vec{n}$ through point $ P$ . Let $ X$ be the perpendicular projection of a given point $ Q$ onto this plane. The so called perpendicular vector of point $ Q$ onto the plane is given by

$\displaystyle \overrightarrow{XQ} = \frac{\overrightarrow{PQ}\cdot\vec{n}} {\vert\vec{n}\vert^2}\,\vec{n}\,. $

Its length

$\displaystyle d = \frac{\vert\overrightarrow{PQ}\cdot\vec{n}\vert} {\vert\vec{n}\vert} $

is the distance of the plane from $ Q$ .
\includegraphics[width=10cm]{abstand}

Given the plane with normal vector $ \vec{n}=\begin{pmatrix}2\\ 1\\ 2\\ \end{pmatrix}$ through point $ P=(1,2,3)$ . We want to compute the distance $ d$ of point $ Q=(3,2,3)$ from the plane and the foot $ X$ of the perpendicular through this point onto the plane.


First of all we have

$\displaystyle \overrightarrow{PQ} = \begin{pmatrix}2 \\ 0 \\ 0 \end{pmatrix},\quad \overrightarrow{PQ} \cdot \vec{n} = 4,\quad \vert\vec{n}\vert = 3\, . $

So we obtain

$\displaystyle \overrightarrow{XQ} = \frac{4}{9} \begin{pmatrix}2\\ 1\\ 2\\ \end{pmatrix},\quad d = \frac{4}{3} $

and finally we compute

$\displaystyle \vec{x} = \vec{q} - \overrightarrow{XQ} = \frac{1}{9} \begin{pmatrix}19\\ 14\\ 19\\ \end{pmatrix}\, . $

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  automatically generated 9/18/2007