Chapter 7

# Chapter 7 - 7 Vectors Exercises 7.1 1(a 6i 12j 2(a 3 3 3(a...

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7 Vectors Exercises 7.1 1. (a) 6 i + 12 j (b) i + 8 j (c) 3 i (d) 65 (e) 3 2. (a) 3 , 3 (b) 3 , 4 (c) 1 , 2 (d) 5 (e) 5 3. (a) 12 , 0 (b) 4 , 5 (c) 4 , 5 (d) 41 (e) 41 4. (a) 1 2 i 1 2 j (b) 2 3 i + 2 3 j (c) 1 3 i j (d) 2 2 / 3 (e) 10 / 3 5. (a) 9 i + 6 j (b) 3 i + 9 j (c) 3 i 5 j (d) 3 10 (e) 34 6. (a) 3 , 9 (b) 4 , 12 (c) 6 , 18 (d) 4 10 (e) 6 10 7. (a) 6 i + 27 j (b) 0 (c) 4 i + 18 j (d) 0 (e) 2 85 8. (a) 21 , 30 (b) 8 , 12 (c) 6 , 8 (d) 4 13 (e) 10 9. (a) 4 , 12 − − 2 , 2 = 6 , 14 (b) 3 , 9 − − 5 , 5 = 2 , 4 10. (a) (4 i + 4 j ) (6 i 4 j ) = 2 i + 8 j (b) ( 3 i 3 j ) (15 i 10 j ) = 18 i + 7 j 11. (a) (4 i 4 j ) ( 6 i + 8 j ) = 10 i 12 j (b) ( 3 i + 3 j ) ( 15 i + 20 j ) = 12 i 17 j 12. (a) 8 , 0 0 , 6 = 8 , 6 (b) 6 , 0 0 , 15 = 6 , 15 13. (a) 16 , 40 − − 4 , 12 = 20 , 52 (b) 12 , 30 − − 10 , 30 = 2 , 0 14. (a) 8 , 12 10 , 6 = 2 , 6 (b) 6 , 9 25 , 15 = 31 , 24 15. −−−→ P 1 P 2 = 2 , 5 16. −−−→ P 1 P 2 = 6 , 4 17. −−−→ P 1 P 2 = 2 , 2 18. −−−→ P 1 P 2 = 2 , 3 290

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Exercises 7.1 19. Since −−−→ P 1 P 2 = −−→ OP 2 −−→ OP 1 , −−→ OP 2 = −−−→ P 1 P 2 + −−→ OP 1 = (4 i + 8 j ) + ( 3 i + 10 j ) = i + 18 j , and the terminal point is (1 , 18). 20. Since −−−→ P 1 P 2 = −−→ OP 2 −−→ OP 1 , −−→ OP 1 = −−→ OP 2 −−−→ P 1 P 2 = 4 , 7 − − 5 , 1 = 9 , 8 , and the initial point is (9 , 8). 21. a (= a ), b (= 1 4 a ), c (= 5 2 a ), e (= 2 a ), and f (= 1 2 a ) are parallel to a . 22. We want 3 b = a , so c = 3(9) = 27. 23. 6 , 15 24. 5 , 2 25. a = 4 + 4 = 2 2; (a) u = 1 2 2 2 , 2 = 1 2 , 1 2 ; (b) u = 1 2 , 1 2 26. a = 9 + 16 = 5; (a) u = 1 5 3 , 4 = 3 5 , 4 5 ; (b) u = 3 5 , 4 5 27. a = 5; (a) u = 1 5 0 , 5 = 0 , 1 ; (b) u = 0 , 1 28. a = 1 + 3 = 2; (a) u = 1 2 1 , 3 = 1 2 , 3 2 ; (b) u = 1 2 , 3 2 29. a + b = 5 , 12 = 25 + 144 = 13; u = 1 13 5 , 12 = 5 13 , 12 13 30. a + b = 5 , 4 = 25 + 16 = 41; u = 1 41 5 , 4 = 5 41 , 4 41 31. a = 9 + 49 = 58; b = 2( 1 58 )(3 i + 7 j ) = 6 58 i + 14 58 j 32. a = 1 4 + 1 4 = 1 2 ; b = 3( 1 1 / 2 )( 1 2 i 1 2 j ) = 3 2 2 i 3 2 2 j 33. 3 4 a = 3 , 15 / 2 34. 5( a + b ) = 5 0 , 1 = 0 , 5 35. 36. 37. x = ( a + b ) = a b 38. x = 2( a b ) = 2 a 2 b 39. b = ( c ) a ; ( b + c ) + a = 0 ; a + b + c = 0 40. From Problem 39, e + c + d = 0 . But b = e a and e = a + b , so ( a + b ) + c + d = 0 . 41. From 2 i + 3 j = k 1 b + k 2 c = k 1 ( i + j ) + k 2 ( i j ) = ( k 1 + k 2 ) i + ( k 1 k 2 ) j we obtain the system of equations k 1 + k 2 = 2, k 1 k 2 = 3. Solving, we find k 1 = 5 2 and k 2 = 1 2 . Then a = 5 2 b 1 2 c . 42. From 2 i + 3 j = k 1 b + k 2 c = k 1 ( 2 i + 4 j ) + k 2 (5 i + 7 j ) = ( 2 k 1 + 5 k 2 ) i + (4 k 1 + 7 k 2 ) j we obtain the system of equations 2 k 1 + 5 k 2 = 2, 4 k 1 + 7 k 2 = 3. Solving, we find k 1 = 1 34 and k 2 = 7 17 . 43. From y = 1 2 x we see that the slope of the tangent line at (2 , 2) is 1. A vector with slope 1 is i + j . A unit vector is ( i + j ) / i + j = ( i + j ) / 2 = 1 2 i + 1 2 j . Another unit vector tangent to the curve is 1 2 i 1 2 j . 44. From y = 2 x + 3 we see that the slope of the tangent line at (0 , 0) is 3. A vector with slope 3 is i + 3 j . A unit vector is ( i + 3 j ) / i + 3 j = ( i + 3 j ) / 10 = 1 10 i + 1 10 j . Another unit vector is 1 10 i 1 10 j .
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