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Homework Statement
The three lines intersect in the point (1; 1; 1): (1 - t; 1 + 2*t; 1 + t), (u; 2*u - 1; 3*u - 2), and (v - 1; 2*v - 3; 3 - v). How can I find three planes which also intersect in the point (1; 1; 1) such that each plane contains one and only one of the three lines?
Homework Equations
plane aix + biy + ciz = di
The Attempt at a Solution
I get 9 equations:Sharing equations with the lines:
a1(1 - t) + b1(1 + 2*t) + c1(1 + t) = d1
a2(u) + b2(2*u - 1) + c2(3*u - 2) = d2
a3(v - 1) + b3(2*v - 3) + c3(3 - v) = d3
Intersection at (1,1,1):
a1 + b1 + c1 = d1
a2 + b2 + c2 = d2
a3 + b3 + c3 = d3
Dot product of plane normals and line vectors = 0 since perpendicular:
<a1; b1; c1> dot <-1; 2; 1> = -a1 + 2*b1 + c1 = 0
<a2; b2; c2> dot <1; 2; 3> = a2 + 2*b2 + 3*c2 = 0
<a3; b3; c3> dot <1; 2; -1> = a3 + 2*b3 - c3 = 0
I know how to find the intersection of 3 planes using matrices/row reduction, and I know some relationships between lines and planes. However, I seem to come up with 12 unknowns and 9 equations for this problem. I know the vectors for the lines must be perpendicular to the normals of the planes, thus the dot product between the two should be 0. I also know that the planes pass through the point (1,1,1) and the x,y,z coordinates for the parameters given in the line equations. What information am I missing? Maybe there are multiple solutions. If so, how can these planes be described with only a line and one point? Another thought was to convert the planes to parametric form, but to describe a plane with parameters normally I would have 2 vectors and one point, but here I only have one vector and one point.