Notes: If (2a+1)u = (b+1)v + (3c-2)w, what constrains a,b and c if w,u and v are not coplanar?

Saturday, June 1, 2013

Since the planes u, v and w are not coplanar they do not
all intersect at any one point (or line, or plane).

This
means that when u and v intersect, there is no point on the plane w that lies on this
line.

Suppose that u and v intersect on the plane z = r,
and values of z on w are denoted z(w) then

(3c - 2)z(w) =
r(2a+1-b-1) = r(2a-b)

Since z(w) is never equal to r, this
implies that a,b and c are defined by the fact that

$\displaystyle{2}{a}-{b}$
!= 3c-2 $\displaystyle{\quad\text{or}\quad}$ 2a-b-3c != -2

So, if $\displaystyle{b}={m}{a}$ then $\displaystyle{c}$
cannot equal $\displaystyle\frac{{{\left({2}-{m}\right)}{a}+{2}}}{{3}}$ .

Therefore
a, b, c can lie anywhere in $\displaystyle{\mathbb{R}}^{{3}}$ except on the
line

(x,y,z) = (0,0,2/3) +
t(1,m,(2-m)/3)

where m =
b/a

Notes