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Unformatted text preview: Assignment 8 Math 417 â€” Winter 2009 Due March 20 Â§ 5.1#40. Weâ€™re given that (among other things) ~v 2 Â· ~v 2 = a 22 = 9. Since ~v 2 Â· ~v 2 = k ~v 2 k 2 , we conclude that k ~v 2 k = 3 (not 3, because lengths are never negative). Â§ 5.1#42. Using the fact that k ~x k 2 = ~x Â· ~x for every vector ~x , we have k ~v 1 + ~v 2 k 2 = ( ~v 1 + ~v 2 ) Â· ( ~v 1 + ~v 2 ) = ~v 1 Â· ~v 1 + 2( ~v 1 Â· ~v 2 ) + ~v 2 Â· ~v 2 = a 11 + 2 a 12 + a 22 = 3 + 10 + 9 = 22 , and so k ~v 1 + ~v 2 k = âˆš 22. Â§ 5.1#44. The vector we seek, being in the span of ~v 2 and ~v 3 , must be a linear combination p~v 2 + q~v 3 for some coefficients p and q ; our task is to find the coefficients so that this combination will be orthogonal to ~v 3 . That is, we want 0 = ( p~v 2 + q~v 3 ) Â· ~v 3 = p ( ~v 2 Â· ~v 3 ) + q ( ~v 3 Â· ~v 3 ) = pa 23 + qa 33 = 20 p + 49 q. There are infinitely many solutions; avoiding fractions, Iâ€™ll take p = 49 and q = 20. (But any multiple of this pair would work as well.) Â§ 5.1#46. Weâ€™re asked to find the coefficients p and q such that the orthogonal projection of ~v 3 to the space spanned by ~v 1 and ~v 2 is p~v 1 + q~v 2 . Since ~v 1 Â· ~v 2 6 = 0 we cannot use the projection formulas. We could apply the Gramâ€“Schmidt algorithm to get an orthonormal basis out of ~v 1 and ~v 2 and then apply the projection formula, but here is a more direct approach. By definition of orthogonal projection, we know that p~v 1 + q~v 2 is the orthogonal projection of ~v 3 onto V exactly when ~v 3 ( p~v 1 + q~v 2 ) is orthogonal to V , i.e., is orthogonal to, i....
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This note was uploaded on 10/13/2009 for the course MATH 417 taught by Professor Eveet during the Spring '09 term at University of MichiganDearborn.
 Spring '09
 eveet
 Math, Linear Algebra, Algebra

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