12-5 - Solutions to Homework Assignment 5 MATH 249-01 and...

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Solutions to Homework Assignment 5 MATH 249-01 and -02 Section 12.5, Page 829 1, 2, 3, 6, 9, 11, 13, 14, 19-24, 26, 30, 37, 39, 40, 45-50, 53, 54, 58, 65-69 3. A vector equation should look like r ( t ) = < - 2 , 4 , 10 > + t < 3 , 1 , - 8 > . Comparing components gives us the parametric equations x = - 2 + 3 t,y = 4 + t,z = 10 - 8 t. 6. A direction vector is just < 1 , 2 , 3 >, so our equations are x = t,y = 2 t,z = 3 t. Symmetric equations are x = y 2 = z 3 . 9. A direction vector is < 2 - 0 , 1 - 1 / 2 , - 3 - 1 > = < 2 , 1 / 2 , - 4 > . Parametric equations are then x = 2+2 t,y = 1+ t/ 2 ,z = - 3 - 4 t. Solving for t gives the symmetric equations x - 2 2 = 2 y - 2 = z + 3 - 4 . 11. The given line has parametric equations x = t - 2 ,y = 2 t,z = t + 3 . Its direction vector is thus < 1 , 2 , 1 > (from the coefficients of t ). We get x = 1 + t,y = - 1 + 2 t,z = 1 + t. 19. L 1 is in the direction < - 6 , 9 , - 3 >, and L 2 is in the direction < 2 , - 3 , 1 > . These are parallel, so the lines are parallel. 21.
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This note was uploaded on 04/07/2008 for the course MTH 249 taught by Professor Starr during the Spring '08 term at Willamette.

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12-5 - Solutions to Homework Assignment 5 MATH 249-01 and...

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