to Solution HW1 Section 6.1 4. F + = ( 2 + 8)i + j 4k; F G = ( 2 8)i + j 8k; F = 39; G G = 68; 2F = 2 2 + 2j 12; 3G = 24i + 6k. 9. F + G = 2i 5j; F G = j.
1
3
G
0
2
1 F=i2j 2 G=i3j 3
1
FG
0
1
F=i2j
4 F+G 5 1 0 1 2 3
2
G=i3j
3 2 1 0 1 2
19. Parametric equations are x = 0, y = 1 + t, z = 3 + 2t; < t < ; Normal equations are x = 0, y 1 = z3 . 2 Section 6.2 4. F G = 53; cos() = 5698 = F G . 10. 3x + 2y = 1
53 ; 77 74
not orthogonal;|F G| = 53 =
2809 <
Section 6.3 10. not colinear; 3x 8y + 2z = 4 3 0 10 6 8 = 34. 19. Volume= 1 3 1 10 Section 6.4 1 6. F + G = 4e1 + 3e2 + 3e3 15e4 e5 ; F G = 1; cos() = 146116 . 12. S is a subspace of R7 . Clearly 0 is in S. Also for F, G in S, F + G and will G have third and fth components equal to zero. 1
13. S is a subspace of R4 since 0, F + G and G will all have equal rst and second components whenever F and G in S have. Section 6.5 9. Suppose 1 (2, 0, 0, 1, 1)+2 (1, 0, 0, 0, 0)+3 (0, 0, 0, 0, 2)+4 (1, 1, 3, 3, 1) = 0. Then certainly 4 = 0 and 21 + 2 + 4 = 0, 1 + 34 = 0 and 1 + 23 + 4 = 0. The only solution of these equations is 1 = 2 = 3 = 4 = 0. Thus the vectors are linearly independent. 860 15. [8i + 6j, 2i 4j, i + k] = 2 4 0 = 44 = 0, so the vectors are linearly 101 independent. 18. A basis for S is the set of vectors (1, 0, 2, 0) and (0, 1, 0, 3), so S has dimension two. 22. A basis for S is the set of vectors (1, 1, 0, 0) and (0, 0, 1, 2), so S has dimension two.
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