User talk:Paul Wormer/scratchbook1

PD Image
Equation for plane. X is arbitary point in plane; $\scriptstyle {\vec {\mathbf {a} }}$ and $\scriptstyle {\hat {\mathbf {n} }}$ 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

${\overrightarrow {OA}}\equiv {\vec {\mathbf {a} }}$

is orthogonal to the plane. The collinear vector

${\hat {\mathbf {n} }}\equiv {\frac {\vec {\mathbf {a} }}{a}}\quad {\hbox{with}}\quad a\equiv {|{\vec {\mathbf {a} }}|}$

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

$\left({\vec {\mathbf {r} }}-{\vec {\mathbf {a} }}\right)\cdot {\hat {\mathbf {n} }}=0\quad {\hbox{with}}\quad {\vec {\mathbf {r} }}\equiv {\overrightarrow {OX}}\quad {\hbox{and}}\quad {\overrightarrow {AX}}={\vec {\mathbf {r} }}-{\vec {\mathbf {a} }}.$

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

$\mathbf {r} \cdot \mathbf {n} =\mathbf {a} \cdot \mathbf {n} \Longleftrightarrow xa_{x}+ya_{y}+za_{z}=a$

with

$\mathbf {a} =(a_{x},\;a_{y},\;a_{z}),\quad \mathbf {r} =(x,\;y,\;z),\quad {\hbox{and}}\quad \mathbf {a} \cdot \mathbf {n} =\mathbf {a} \cdot {\frac {\mathbf {a} }{a}}=a={\sqrt {a_{x}^{2}+a_{y}^{2}+a_{z}^{2}}}.$

Conversely, given the following equation for a plane

$ax+by+cz=d\,$

writing

$\mathbf {r} =(x,\;y,\;z),\quad \mathbf {f} =(a,\;b,\;c),\quad {\hbox{and}}\quad \mathbf {a} \equiv {\frac {d}{\mathbf {f} \cdot \mathbf {f} }}\mathbf {f}$

it follows that

$\mathbf {f} \cdot \mathbf {r} =d=\mathbf {f} \cdot \mathbf {a} .$

Hence

$\mathbf {f} \cdot (\mathbf {r} -\mathbf {a} )=0\;\Longrightarrow \;\mathbf {n} \cdot (\mathbf {r} -\mathbf {a} )=0\quad {\hbox{with}}\quad \mathbf {n} ={\frac {\mathbf {f} }{\sqrt {a^{2}+b^{2}+c^{2}}}}$

where f , a, and n are collinear.