# 4.2 - Math 2940 Solutions Fall 2011 Section 4 2 2(a ~a T ~a...

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Unformatted text preview: Math 2940 Solutions, Fall 2011 Section 4 . 2 2 ) (a) ~a T ~a = 1 and ~a T ~ b = cos so our projection is cos . (b) ~a T ~a = 2 and ~a T ~ b = 0 so our projection is the zero vector! 5 ) For ~a 1 = (- 1 , 2 , 2) we have ~a T 1 ~a 1 = 9 and - 1 2 2 - 1 2 2 = 1- 2- 2- 2 4 4- 2 4 4 so the projection matrix in this case is 1 / 9- 2 / 9- 2 / 9- 2 / 9 4 / 9 4 / 9- 2 / 9 4 / 9 4 / 9 . For ~a 2 = (2 , 2 ,- 1) we have ~a T 1 ~a 1 = 9 again and 2 2- 1 2 2- 1 = 4 4- 2 4 4- 2- 2- 2 1 so the projection matrix in this case is 4 / 9 4 / 9- 2 / 9 4 / 9 4 / 9- 2 / 9- 2 / 9- 2 / 9 1 / 9 . Finally 4 / 9 4 / 9- 2 / 9 4 / 9 4 / 9- 2 / 9- 2 / 9- 2 / 9 1 / 9 1 / 9- 2 / 9- 2 / 9- 2 / 9 4 / 9 4 / 9- 2 / 9 4 / 9 4 / 9 = 0 0 0 0 0 0 0 0 0 . Why do we get the matrix? Well, the dot product of ~a 1 and ~a 2 is 0, so the lines through these vectors are perpendicular. So if we project onto one of them first, the subsequent projectionvectors are perpendicular....
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## This note was uploaded on 01/02/2012 for the course MATH 2940 at Cornell.

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4.2 - Math 2940 Solutions Fall 2011 Section 4 2 2(a ~a T ~a...

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