hw13.solutions

X x u2 x u3 4 1 1 x u1 u1 u2 u3 u1 u2 u3 u1 u1 u2

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Unformatted text preview: u1 + u2 = − u1 + u2 u1 · u1 u2 · u2 2 4 #10. x= x · u2 x · u3 4 1 1 x · u1 u1 + u2 u3 = u1 + u2 + u3 u1 · u1 u2 · u2 u3 · u3 3 3 3 #12. Take u = (−1, 3), so the projection is y·u 2 u = − u = (2/5, −6/5) u·u 5 #16. The projection is 15 y·u u= (1, 2) = (3, 6) u·u 5 so the distance is ||(−3, 9) − (3, 6)|| = ||(−6, −3)|| = √ 36 + 9 = √ 45. v u1 Section 6.3, #2. The projection on the line through u1 is w = u1··u1 u1 = 2u1 = (2, 4, 2, 2). This is the in the span of u1 , and the difference v − 2u1 = (2, 1, −5, 1) is the span of u2...
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This note was uploaded on 12/15/2013 for the course MATH 461 taught by Professor Jake during the Fall '08 term at Maryland.

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