muckymotter Posted May 8, 2007 Posted May 8, 2007 Consider the quadrilateral (namely Q) in R^3 formed by the points (1, 0, 0), (2, 0, 0), (1, 1, 3), and (2, 1, 3). (a) What should the coordinates be for the figure R we get by rotating Q counterclockwise in the x-y plane by 45 degrees, then dilating it by a factor of 3/2, then translating it along the vector (-2, 1, -1)? Okay, so what I did was I used the matrix cos45 -sin45 0 sin45 cos45 0 0 0 1 After this, multiplied the identity matrix for R^3 by 3/2 and then multiplied it by the matrix with 45 substituted for theta. Then T(x,y,z)=(-2+3sqrt(2)x/4-3sqrt(2)y/4,1+3sqrt(2)x/4+3sqrt(2)y/4,-1+3z/2) I substituted each of the quadrilateral points in for T(x, y, z) to come up with the four points and got: (3/(2sqrt2) - 2, 3/(2sqrt2) + 1, -1), (3/sqrt2 - 2, 3/sqrt2 + 1, -1), (-2, 3/sqrt2 + 1, 3.5), (3/(2sqrt2) - 2, 9/(2sqrt2) + 1, 3.5) But my trouble lies in doing (b) so if anyone could explain how it is done, I would be really grateful as I have been stuck on this for days!: (b) What matrix transforms Q into R?
Bignose Posted May 8, 2007 Posted May 8, 2007 Assuming that that answer for a is right (I didn't check, and I assume you have checked), you've almost done the entire answer. Let me call the first matrix Q and second matrix (your answer) R. These are related by Q = H*R where here H is the transformation matrix. Well, to isolate H, you just right multiply both sides of the this equation by the inverse of R, R^-1. Since a matrix times its inverse is the identity matrix on the right hand side of the equation you get H times the identity matrix. Then, any matrix times the identity matrix is itself, so the right hand side is just H. The equation becomes Q*R^-1 = H. So, you just need to get R^-1 and multiply Q by R^-1 to get H, the transformation matrix you seek.
muckymotter Posted May 8, 2007 Author Posted May 8, 2007 Thanks! I'm a little confused, though - The first matrix Q is a 3x3 matrix, but the transformation matrix R is a 4x3 so I can't multiply them.
Bignose Posted May 9, 2007 Posted May 9, 2007 Well, there is a generalized matrix inverse also known as pseudoinverse for non-square matrices. The pseduoinversed matrix R would be a 3x4 matrix, and you can right mulitply a 3x3 matrix by a 3x4 matrix.
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