hwsolB14

# hwsolB14 - 2\sqrt{2}*c2 = -3\sqrt{2} 4*c1 + 1*c2 = b The...

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Sheet1 Page 1 HW 14 Solution ______________ (i) ||v1|| = ||v2|| = sqrt(1 + 8 + 16) = 5 <v1,v2> = 4-8+4 = 0 (ii) <v1,a> = 3+12+12 = 27, <v2,a> = 12-12+3 = 3 ||a|| = sqrt(9 + 18 + 9) = sqrt(36) = 6 f1 = ( <v1,a> / ||v1||^2 )*v1 = (27/25)*v1 Angle between a and v1 = arccos(27/(6*5)) = 0.4510 rad f2 = ( <v2,a> / ||v2||^2 )*v2 = (3/25)*v2 Angle between a and v1 = arccos(3/(6*5)) = 1.4706 rad (iii) If [3 -4 scalars c1 and c2, then c1 + 4*c2 = 3 -2\sqrt(2)*c1 +

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Unformatted text preview: 2\sqrt{2}*c2 = -3\sqrt{2} 4*c1 + 1*c2 = b The first two equations have the unique solution c1 = 9/5, c2 = 3/10 which means that b = 7.5. (iv) Since v1 and v2 are orthogonal, the projection of a on the subspace (plane) generated by v1 and v2 will be the sum of the projections on v1 and v2, i.e., (27/25)*v1 + (3/25)*v2 = (1/25)*[ 39 -48*sqrt(2) 111 ].' Sheet1 Page 2 b] can be expressed as c1*v1 + c2*v2 for some...
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## This note was uploaded on 03/21/2010 for the course ENEE 241 taught by Professor Staff during the Spring '08 term at Maryland.

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hwsolB14 - 2\sqrt{2}*c2 = -3\sqrt{2} 4*c1 + 1*c2 = b The...

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