AEM_3e_Chapter_07

AEM_3e_Chapter_07 - Part II Vectors, Matrices, and Vector...

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Part II Vectors, Matrices, and Vector Calculus 7 7 Vectors EXERCISES 7.1 Vectors in 2-Space 3. (a) h 12 , 0 i (b) h 4 , 5 i (c) h 4 , 5 i (d) 41 (e) 41 6. (a) h 3 , 9 i (b) h− 4 , 12 i (c) h 6 , 18 i (d) 4 10 (e) 6 10 9. (a) h 4 , 12 i−h− 2 , 2 i = h 6 , 14 i (b) h− 3 , 9 5 , 5 i = h 2 , 4 i 12. (a) h 8 , 0 i−h 0 , 6 i = h 8 , 6 i (b) h− 6 , 0 0 , 15 i = h− 6 , 15 i 15. −−−→ P 1 P 2 = h 2 , 5 i 18. P 1 P 2 = h 2 , 3 i 21. a (= a ), b (= 1 4 a ), c (= 5 2 a ), e (= 2 a ), and f (= 1 2 a ) are parallel to a . 24. h 5 , 2 i 27. k a k =5; (a) u = 1 5 h 0 , 5 i = h 0 , 1 i ; (b) u = h 0 , 1 i 30. k 2 a 3 b k = kh− 5 , 4 ik = 25+16= 41; u = 1 41 h− 5 , 4 i = h− 5 41 , 4 41 i 33. 3 4 a = h− 3 , 15 / 2 i 36. 39. b =( c ) a ;( b + c )+ a = 0 ; a + b + c = 0 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 k 2 ) j we obtain the system of equations 2 k 1 k 2 =2,4 k 1 k 2 = 3. Solving, we ±nd k 1 = 1 34 and k 2 = 7 17 . 101
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7.1 Vectors in 2-Space 45. (a) Since F f = F g , k F g k = k F f k = µ k F n k and tan θ = k F g k / k F n k = µ k F n k / k F n k = µ . (b) θ = tan 1 0 . 6 31 48. Place one corner of the parallelogram at the origin and let two adja- cent sides be −−→ OP 1 and 2 . Let M be the midpoint of the diagonal connecting P 1 and P 2 and N be the midpoint of the other diagonal. Then OM = 1 2 ( 1 + 2 ). Since 1 + 2 is the main diagonal of the parallelogram and N is its midpoint, ON = 1 2 ( 1 + 2 ). Thus, = and the diagonals bisect each other. EXERCISES 7.2 Vectors in 3-Space 3. 6. 9. A line perpendicular to the xy -plane at (2 , 3 , 0) 12. 15. The union of the planes x =0, y = 0, and z =0 18. The union of the planes x = 2 and z =8 21. d = p (3 6) 2 +( 1 4) 2 +(2 8) 2 = 70 24. (a) 2; (b) d = p ( 6) 2 +2 2 3) 2 =7 27. d ( P 1 ,P 2 )= p (4 1) 2 +(1 2) 2 +(3 3) 2 = 10 d ( P 1 3 p (4 1) 2 +(6 2) 2 +(4 3) 2 = 26 d ( P 2 3 p (4 4) 2 1) 2 3) 2 = 26; The triangle is an isosceles triangle. 30. d ( P 1 2 p (1 2) 2 3) 2 2) 2 = 6 102
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7.3 Dot Product d ( P 1 ,P 3 )= p (5 2) 2 +(0 3) 2 +( 4 2) 2 =3 6 d ( P 2 3 p (5 1) 2 4) 2 4 4) 2 =4 6 Since d ( P 1 2 )+ d ( P 1 3 d ( P 2 3 ), the points P 1 , P 2 , and P 3 are collinear. 33. µ 1+7 2 , 3+( 2) 2 , 1 / 2+5 / 2 2 =(4 , 1 / 2 , 3 / 2) 36. ( 5)) / 2= x 3 = 4; (4 + 8) / y 3 =6; (1+3) / z 3 =2. The coordinates of P 3 are ( 4 , 6 , 2). (a) µ 4) 2 , 4+6 2 , 1+2 2 =( 7 / 2 , 5 , 3 / 2) (b) µ 4+( 5) 2 , 6+8 2 , 2+3 2 9 / 2 , 7 , 5 / 2) 39. −−−→ P 1 P 2 = h 2 , 1 , 1 i 42. 2 a ( b c h 2 , 6 , 4 i−h− 3 , 5 , 8 i = h 5 , 1 , 12 i 45. k a + c k = kh 3 , 3 , 11 ik = 9+9+121= 139 48. k b k a + k a k b = 1+1+1 h 1 , 3 , 2 i + 1+9+4 h− 1 , 1 , 1 i = h 3 , 3 3 , 2 3 i + h− 14 , 14 , 14 i = h 3 14 , 3 3+ 14 , 2 14 i 51. b a i 4 j +4 k EXERCISES 7.3 Dot Product 3. a · b =2( 1)+( 3)2 + 4(5) = 12 6. a · ( b + c ) = 2(2) + ( 3)8 + 4(4) = 4 9. a · a =2 2 3) 2 2 =29 12. (2 a ) · ( a 2 b ) = 4(4) + ( 6)( 7)+8( 6) = 10 15. a and f, b and e, c and d 18. If a and b represent adjacent sides of the rhombus, then k a k = k b k , the diagonals of the rhombus are a + b and a b , and ( a + b ) · ( a b a · a a · b + b · a b · b = a · a b · b = k a k 2 −k b k 2 =0 .
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AEM_3e_Chapter_07 - Part II Vectors, Matrices, and Vector...

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