User talk:Paul Wormer/scratchbook1

Point-normal representation

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Fig. 1. Equation for plane. P is arbitary point in plane; $\scriptstyle {\vec {d}}$ and $\scriptstyle {\vec {n}}_{0}$ 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 figure 1. Point P 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

${\vec {d}}\equiv {\overrightarrow {OA}}$

is orthogonal to the plane. The collinear vector

${\vec {n}}_{0}\equiv {\frac {1}{d}}{\vec {d}}\quad {\hbox{with}}\quad d\equiv \left|{\vec {d}}\,\right|$

is a unit (length 1) vector normal (perpendicular) to the plane which is known as the normal of the plane in point A. Note that d is the distance of O to the plane. The following relation holds for an arbitrary point P in the plane

$\left({\vec {r}}-{\vec {d}}\;\right)\cdot {\vec {n}}_{0}=0\quad {\hbox{with}}\quad {\vec {r}}\equiv {\overrightarrow {OP}}\quad {\hbox{and}}\quad {\vec {r}}-{\vec {d}}={\overrightarrow {AP}}.$

This equation for the plane can be rewritten in terms of coordinates with respect to a Cartesian frame with origin in O. Dropping arrows for component vectors (real triplets) that are written bold, we find

$\left(\mathbf {r} -\mathbf {d} \right)\cdot \mathbf {n} _{0}=0\Longleftrightarrow xa_{0}+yb_{0}+zc_{0}=d$

with

$\mathbf {d} =(a,\;b,\;c),\quad \mathbf {n} _{0}=(a_{0},\;b_{0},\;c_{0}),\quad \mathbf {r} =(x,\;y,\;z),$

and

$\mathbf {d} \cdot \mathbf {n} _{0}={\frac {1}{d}}\mathbf {d} \cdot \mathbf {d} =d={\sqrt {a^{2}+b^{2}+c^{2}}}.$

Conversely, given the following equation for a plane

$ax+by+cz=e,\,$

it is easy to derive the same equation. Write

$\mathbf {r} =(x,\;y,\;z),\quad \mathbf {f} =(a,\;b,\;c),\quad {\hbox{and}}\quad \mathbf {d} \equiv \left({\frac {e}{a^{2}+b^{2}+c^{2}}}\right)\mathbf {f} .$

It follows that

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

Hence we find the same equation,

$\mathbf {f} \cdot (\mathbf {r} -\mathbf {d} )=0\;\Longrightarrow \;(\mathbf {r} -\mathbf {d} )\cdot \mathbf {n} _{0}=0\quad {\hbox{with}}\quad \mathbf {n} _{0}={\frac {1}{\sqrt {a^{2}+b^{2}+c^{2}}}}\mathbf {f}$

where f , d, and n₀ are collinear. The equation may also be written in the following mnemonically convenient form

$\mathbf {d} \cdot (\mathbf {r} -\mathbf {d} )=0,$

which is the equation for a plane through a point A perpendicular to ${\overrightarrow {OA}}$.

Three-point representation

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Fig. 2. Plane through points A, B, and C.

Figure 2 shows a plane that by definition passes through non-coinciding points A, B, and C that moreover are not on one line. The point P is an arbitrary point in the plane and the reference point O is outside the plane. Referring to figure 2 we introduce the following definitions

${\vec {a}}={\overrightarrow {OA}},\quad {\vec {b}}={\overrightarrow {OB}},\quad {\vec {c}}={\overrightarrow {OC}},\quad {\vec {r}}={\overrightarrow {OP}}.$

Clearly the following two non-collinear vectors belong to the plane

${\vec {u}}={\overrightarrow {AB}}={\vec {b}}-{\vec {a}},\quad {\vec {v}}={\overrightarrow {AC}}={\vec {c}}-{\vec {a}}.$

Because a plane (an affine space), with a given fixed point as origin is a 2-dimensional linear space and two non-collinear vectors with "tails" in the origin are linearly independent, it follows that any vector in the plane can be written as a linear combination of these two non-collinear vectors (this is also expressed as: any vector in the plane can be decomposed into components along the two non-collinear vectors). In particular, taking A as origin in the plane,

${\overrightarrow {AP}}={\vec {r}}-{\vec {a}}=\lambda {\vec {u}}+\mu {\vec {v}},\qquad \lambda ,\mu \in \mathbb {R} .$

The real numbers λ and μ specify the direction of ${\overrightarrow {AP}}$. Hence the following equation for the position vector ${\vec {r}}$ of the arbitrary point P in the plane:

${\vec {r}}={\vec {a}}+\lambda {\vec {u}}+\mu {\vec {v}}$

is known as the point-direction representation of the plane. This representation is equal to the three-point representation

${\vec {r}}={\vec {a}}+\lambda ({\vec {b}}-{\vec {a}})+\mu ({\vec {c}}-{\vec {a}}),$

where ${\vec {a}}$, ${\vec {b}}$, and ${\vec {c}}$ are the position vectors of the three points that define the plane.

Writing for the position vector of the arbitrary point P in the plane

${\vec {r}}=(1-\lambda -\mu )\,{\vec {a}}+\lambda \,{\vec {b}}+\mu \,{\vec {c}}\;\equiv \;\xi _{1}\,{\vec {a}}+\xi _{2}\,{\vec {b}}+\xi _{3}\,{\vec {c}}\;,$

we find that the real triplet (ξ₁, ξ₂, ξ₃) with ξ₁ + ξ₁ + ξ₁ = 1 forms a set of coordinates for P. The numbers {ξ₁, ξ₂, ξ₃ | ξ₁+ ξ₂+ ξ₃ = 1 } are known as the barycentric coordinates of P. It is trivial to go from barycentric coordinates to the "three-point representation",

${\vec {r}}=\xi _{1}{\vec {a}}+\xi _{2}{\vec {b}}+\xi _{3}{\vec {c}}\quad {\hbox{with}}\quad \xi _{1}=1-\xi _{2}-\xi _{3}\;\Longleftrightarrow \;{\vec {r}}={\vec {a}}+\xi _{2}({\vec {b}}-{\vec {a}})+\xi _{3}({\vec {c}}-{\vec {a}}).$