Question

[Calculus:Linear systems and planes:Geometry of systems of equations]
Write a 3-by-3 system of equations:

Where the planes are all different and all intersect in a line.
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Is this correct?

https://www.dropbox.com/s/eoldjxrcd6ep24e/IMG_20160124_144439.jpg?dl=0

Answer 1

There are a few ways to think about this problem. If you have all the planes go through the origin (as in your case), and you pick two planes (by defining their "normals" N1 and N2) (so that e.g. N1 dot X = 0 where X is a 3-tuple i.e. point on the plane
then you can create a third normal (to a plane) by any linear combination of the first 2 normals. As long as you don't use 0 as one of the coefficients in
a N1 + b N2 = N3
this third normal will be different from the other two, but will lie in the same plane (whose normal is found by N1 cross N2, if you wanted to do that)

Answer 2

So your posted solution is one of an infinite number of solutions for finding the third plane as a linear combination of the first two planes.

The intersection of all three planes will be a line defined by the "direction vector" which is the normal to the plane in which all 3 normals lie, if that makes any sense.

Answer 3

Thank you very much.