hw13.solutions

hw13.solutions - Math 461/Fall 2013/Jerey Adams Solutions...

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Math 461/Fall 2013/Jefrey Adams Solutions to Homework 13 (originally due 12/11/13 but canceled due to snow) Section 6.2, #2, 4, 6, 8, 10, 12, 16; Section 6.3, #2, 4, 6, 8, 12, 14, 18; Section 6.4, #2, 4, 6, 10, 14, 16; Section 6.5, #2, 4, 6, 8, 10 Section 6.2, #2. (1 , 2 , 1) · (0 , 1 , 2) = 0, (1 , 2 , 1) · ( 5 , 2 , 1) = 0, (0 , 1 , 2) · ( 5 , 2 , 1) = 0 so they are orthogonal. #4. All dot products are 0: orthgonoal. #6. ( 4 , 1 , 3 , 8) · (3 , 3 , 5 , 1) n = 0, so not orthogonal. #8. vu 1 · 2 = 0, so they are orthgonal. vx = · 1 1 · 1 1 + · 2 2 · 2 2 = 3 2 1 + 3 4 2 #10. = · 1 1 · 1 1 + · 2 2 · 2 2 · 3 3 · 3 3 = 4 3 1 + 1 3 2 + 1 3 3 #12. Take = ( 1 , 3), so the projection is v y · · = 2 5 = (2 / 5 , 6 / 5) #16. The projection is v y · · = 15 5 (1 , 2) = (3 , 6) so the distance is || ( 3 , 9) (3 , 6) || = || ( 6 , 3) || = 36 + 9 = 45. Section 6.3, #2. The projection on the line through 1 is vw = vV · 1 1 · 1 1 = 2 1 = (2 , 4 , 2 , 2). This is the in the span of 1 , and the diFerence 2 1 = (2 , 1 , 5 , 1) is the span of 2 , vu 3 4 #6. The projection is v y · 1 1 · 1 1 + v y · 2 2 · 2 2 = 3 2 1 + 5 2 2 = (6 , 4 , 1) #10. The projection is v y · 1 1 · 1 1 + v y ·
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hw13.solutions - Math 461/Fall 2013/Jerey Adams Solutions...

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