< User talk:Paul WormerRevision as of 06:13, 30 March 2010 by imported>Paul Wormer
PD Image Equation for plane. X is arbitary point in plane;
and
are collinear.
In analytic geometry several closely related algebraic equations are known for a plane in three-dimensional Euclidean space. One such equation is illustrated in the figure. Point X is an arbitrary point in the plane and O (the origin) is outside the plane. The point A in the plane is chosen such that vector
![{\displaystyle {\vec {\mathbf {a} }}\equiv {\overrightarrow {OA}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6b7a9128412d4d608dddf70850970df5c53af461)
is orthogonal to the plane. The collinear vector
![{\displaystyle {\hat {\mathbf {n} }}\equiv {\frac {\vec {\mathbf {a} }}{a}}\quad {\hbox{with}}\quad a\equiv {|{\vec {\mathbf {a} }}|}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c031470a67ce838e782161f7b1417115e9831d76)
is a unit (length 1) vector normal (perpendicular) to the plane. Evidently a is the distance of O to the plane. The following relation holds for an arbitrary point X in the plane
![{\displaystyle \left({\vec {\mathbf {r} }}-{\vec {\mathbf {a} }}\right)\cdot {\hat {\mathbf {n} }}=0\quad {\hbox{with}}\quad {\vec {\mathbf {r} }}\equiv {\overrightarrow {OX}}\quad {\hbox{and}}\quad {\vec {\mathbf {r} }}-{\vec {\mathbf {a} }}={\overrightarrow {AX}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/075919adb01cf9d23722291956cd781301b7716c)
This equation for the plane can be rewritten in terms of coordinates with respect to a Cartesian frame with origin in O. Dropping arrows and hat for component vectors (real triples), we find
![{\displaystyle \mathbf {r} \cdot \mathbf {n} =\mathbf {a} \cdot \mathbf {n} \Longleftrightarrow xa_{x}+ya_{y}+za_{z}=a}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fe9efd2936670ae5a760108619253402040b56d3)
with
![{\displaystyle \mathbf {a} =(a_{x},\;a_{y},\;a_{z}),\quad \mathbf {r} =(x,\;y,\;z),\quad {\hbox{and}}\quad \mathbf {a} \cdot \mathbf {n} =\mathbf {a} \cdot {\frac {\mathbf {a} }{a}}=a={\sqrt {a_{x}^{2}+a_{y}^{2}+a_{z}^{2}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/88e83988b2a47b8e194f299dadbdf4e0b693a3e8)
Conversely, given the following equation for a plane
![{\displaystyle ax+by+cz=d\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1f2ecac2f60150e5c228f10e50133b72909d2616)
it is easy to derive the same equation.
Write
![{\displaystyle \mathbf {r} =(x,\;y,\;z),\quad \mathbf {f} =(a,\;b,\;c),\quad {\hbox{and}}\quad \mathbf {a} \equiv {\frac {d}{\mathbf {f} \cdot \mathbf {f} }}\mathbf {f} .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c32e6941b3fa87bf37df8f57c82dd5535e5772c1)
It follows that
![{\displaystyle \mathbf {f} \cdot \mathbf {r} =d=\mathbf {f} \cdot \mathbf {a} .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8dbbe6010e10c5177e46d9746fb1e2d415b7ab7f)
Hence we find the same equation,
![{\displaystyle \mathbf {f} \cdot (\mathbf {r} -\mathbf {a} )=0\;\Longrightarrow \;\mathbf {n} \cdot (\mathbf {r} -\mathbf {a} )=0\quad {\hbox{with}}\quad \mathbf {n} ={\frac {\mathbf {f} }{\sqrt {a^{2}+b^{2}+c^{2}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b5fbc5f9465a49ac454f770bd49915dc42e4152c)
where f , a, and n are collinear.