intersection

2 points define a line. 3 points not on a line define a plane.

in the case of two points x1,y1,z1 and x2,y2,z2, if x1 != x2 the relationship between x and y that defines the line through the two points is y = m[xy]x + b where m[xy] = (y1 - y2) / (x1 - x2) and b[xy] is derived as follows: b[xy] is the value of y where x = 0, so m[xy] = (y1 - b[xy]) / (x1 - 0), m[xy](x1) = y1 - b[xy], m[xy](x1) + b[xy] = y1, and b[xy] = y1 - m[xy]x1. similarly, where z1 != z2 the relationship between z and y that defines the line is y = m[zy]z + b[zy] and m[zy] = (y1-y2) / (z1 - z2) and b[zy] = y1 - m[zy]z1.

My purpose is to derive the point of intersection of a line and a plane. I therefore require the formula for a plane, which I will attempt to derive from the three points x[a],y[a],z[a], x[b],y[b],z[b], and x[c],y[c],z[c].

First there are three special cases. If x[a] = x[b] = x[c] = n the formula for the plane is x = n. If y[a] = y[b] = y[c] = n the formula for the plane is y = n. And if z[a] = z[b] = z[c] = n the formula for the plane is z = n.

Second, there are three other special cases. Where m[xyab] =  (y[a] - y[b]) / (x[a] - x[b]) and m[xybc] and m[xyac] are similarly derived, if m[xyab] = m[xybc] = m[xyac] = n the formula for the plane is y = nx + b where be is derived from n and any of the three points's x and y values. Similarly if m[zyab] = m[zybc] = m[zyac] = n the formula for the plane is y = nz + b and if m[xzab] = m[xzbc] = m[xzac] = n the formula for the plane is z = nx + b.

Then, outside of these six special cases, the general case is such that, first, the plane will intersect the y axis at a certain point, then, starting from that point, which will have the coordinates 0,0,b, and for all other values of x,z, there will be a unique value of y which is on the plane. At a guess, the formula for such a plane will be y = m[xy]x + m[zy]z + b[p].

We now have, given the general case line and the general case plane, y = m[xy]x + b[xy] and z = m[xz]x + b[xz] and y = m[xyp]x + m[zyp]z + b[p]. Let's see where it takes us. We are looking for a point which satisfies all three equations. We can presumably begin by asserting that m[xy]x + b[xy] = m[xyp]x + m[zyp]z + b[p]. And z makes an appearance, so m[xy]x + b[xy] = m[xyp]x + m[zyp](m[xz]x + b[xz]) + b[p]. It at least looks possible we can now solve for x.