Solution1_Fall2008

Solution1_Fall2008 - α 4 = 0 and-2 α 1 α 2 α 4 = 0 α 1...

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Solution to HW1 Section 6.1 4. F + G = ( 2 + 8) i + j - 4 k ; F - G = ( 2 - 8) i + j - 8 k ; k F k = 39; k G k = 68; 2 F = 2 2 + 2 j - 12; 3 G = 24 i + 6 k . 9. F + G = 2 i - 5 j ; F - G = j . -1 0 1 2 3 -5 -4 -3 -2 -1 0 1 F=i-2j G=i-3j F+G -2 -1 0 1 2 -3 -2 -1 0 1 2 3 F-G G=i-3j F=i-2j -G 19. Parametric equations are x = 0 ,y = 1 + t,z = 3 + 2 t ; -∞ < t < ; Normal equations are x = 0 ,y - 1 = z - 3 2 . Section 6.2 4. F · G = - 53; cos( θ ) = - 53 77 74 ; not orthogonal; | F · G | = 53 = 2809 < 5698 = k F kk G k . 10. - 3 x + 2 y = 1 Section 6.3 10. not colinear; - 3 x - 8 y + 2 z = 4 19. Volume= - - 3 0 10 1 6 8 - 3 - 1 10 = 34 . Section 6.4 6. F + G = - 4 e 1 + 3 e 2 + 3 e 3 - 15 e 4 - e 5 ; F · G = - 1;cos( θ ) = - 1 146 116 . 12. S is a subspace of R 7 . Clearly 0 is in S . Also for F,G in S , F + G and αG will have third and fifth components equal to zero. 1
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13. S is a subspace of R 4 since 0 , F + G and αG will all have equal first and second components whenever F and G in S have. Section 6.5 9. Suppose α 1 ( - 2 , 0 , 0 , 1 , 1)+ α 2 (1 , 0 , 0 , 0 , 0)+ α 3 (0 , 0 , 0 , 0 , 2)+ α 4 (1 , - 1 , 3 , 3 , 1) = 0 . Then certainly
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Unformatted text preview: α 4 = 0 and-2 α 1 + α 2 + α 4 = 0, α 1 + 3 α 4 = 0 and α 1 + 2 α 3 + α 4 = 0. The only solution of these equations is α 1 = α 2 = α 3 = α 4 = 0. Thus the vectors are linearly independent. 15. [8 i +6 j, 2 i-4 j,i + k ] = fl fl fl fl fl fl 8 6 2-4 0 1 1 fl fl fl fl fl fl =-44 6 = 0, so the vectors are linearly independent. 18. A basis for S is the set of vectors (1 , , 2 , 0) and (0 , 1 , , 3) , so S has dimension two. 22. A basis for S is the set of vectors (-1 , 1 , , 0) and (0 , , 1 , 2) , so S has dimension two. 2...
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This note was uploaded on 09/19/2008 for the course AME 525 taught by Professor Newton during the Fall '08 term at USC.

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Solution1_Fall2008 - α 4 = 0 and-2 α 1 α 2 α 4 = 0 α 1...

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