FinalTestReviewSpring2011

0 1 0 answer an orthonormal basis under question is 1

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Unformatted text preview: v 0 −1 −1 0 1 0 (12 pts) Solution: First, we find orthogonal vectors ￿ 1 , ￿ 2 , and ￿ 3 , and, at the end of uu u the process, normalize each of them to ￿1 , ￿2 , and ￿3 , respectively. q q q 0 1 1 0 1+0+1+0 0 0 ￿1 ￿ uv We have ￿ 1 = ￿1 , ￿ 2 = ￿2 − ￿ 1 ··￿2 ￿ 1 = u vu v u u u1 −1 − 1+0+1+0 −1 = 0 , and 1 1 0 2 1 0 1 −1 2+0+0+0 0 0+0+0+0 0 −1 ￿1 ￿ uv ￿2 ￿ uv ￿ 3 = ￿3 − ￿ 1 ··￿3 ￿ 1 − ￿ 2 ··￿3 ￿ 2 = u v u u u u1 u u2 0 − 1+0+1+0 −1 − 0+0+0+1 0 = 1 . 0 0 1 0 √ √ √ √ 1 u= u + Since ||￿ 1 || = + 0 + 1 + 0 = 2, ||￿ 2 || 1, and ||￿ 3 || = 1 + 1 + 1 0 = 3, we have u 1 0 1 0 0 −1 u ￿2 u ￿3 u 1 1 q √ ￿1 = ||￿ 1 || = √2 q q ￿1 u u u −1 , ￿2 = ||￿ 2 || = 0 , and ￿3 = ||￿ 3 || = 3 1 . 0 1 0 Answer: An orthonormal basis under question is {￿1 , ￿2 , ￿3 }, where vectors ￿i are as above. qqq q 2 MATH 441.001 Instr. K. Ciesielski Spring 2011 NAME (print): Review for TEST # 4 Solve the followin...
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This document was uploaded on 02/19/2014 for the course MATH 441 at WVU.

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