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If (2a+1)u = (b+1)v + (3c-2)w, what constrains a,b and c if w,u and v are not coplanar?
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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
`2a-b != 3c-2` or `2a-b-3c != -2`
So, if `b = ma` then `c` cannot equal `[(2-m)a + 2]/3` .
Therefore a, b, c can lie anywhere in `RR^3` except on the line
(x,y,z) = (0,0,2/3) + t(1,m,(2-m)/3)
where m = b/a
Posted by mathsworkmusic on April 28, 2013 at 7:19 PM (Answer #1)
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