Solution1_Fall2008

Solution1_Fall2008 - 4 = 0 and-2 1 + 2 + 4 = 0, 1 + 3 4 = 0...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 4 = 0 and-2 1 + 2 + 4 = 0, 1 + 3 4 = 0 and 1 + 2 3 + 4 = 0. The only solution of these equations is 1 = 2 = 3 = 4 = 0. Thus the vectors are linearly independent. 15. [8 i +6 j, 2 i-4 j,i + k ] = 8 6 2-4 0 1 1 =-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...
View Full Document

Page1 / 2

Solution1_Fall2008 - 4 = 0 and-2 1 + 2 + 4 = 0, 1 + 3 4 = 0...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online