Figured it out.
By any chance did you do following:
Do the full QR factorization of the two matrices. Using the third column from each 'Q' matrix, build a new matrix, call it Q'. It is composed of the normals to the planes described by the original two matrices. Do a full QR of Q'. The new QR has a third column of Q that is orthogonal to both of those normal vectors. So it is therefore in the intersection.
Is this correct? or did I make a bad assumption?
It seems that we are all in CSE 6643, which also seems all good according to the syllabus and working together. So, qiaoshiya, this seems like a very reasonable assumption. From your thoughts, and talking with Prof Alben, I went back through Chapter 7. On page 50 the text states "Notice that in the full QR factorization, the columns of q_j for j>n are orthogonal to range(A)." That means that the third column of Q should basically be equivalent to cross(x1, y1), which is one way of identifying a plane (use the plane's normal vector). With this, then using the 2 third columns of the Qs, then the third QR factorization would result in a vector that is perpendicular to both of the first two plane identifying vectors. That is exactly what we're are looking for.
When I first saw this problem I went through (mostly) the exercise of finding the the final vector, since I knew how to do that. Now reading that line from page 50, the two processes seem to be identical.
Other thoughts from anyone?
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