# Hence there do not exist scalars c 1 c 2 and c 3

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Hence, there do not exist scalars c 1 , c 2 , and c 3 which satisfy the given equation. 5. (c) v = [ 3 2 + 4 2 + 0 2 + (–12) 2 ] 1/2 = 169 = 13 6. (a) u + v = (4, 4, 10, 1) = 4 2 + 4 2 + 10 2 + 1 2] 1/2 = 133 (c) –2 u + 2 u = [ (–8) 2 + (–2) 2 + (–4) 2 + (–6) 2 ] 1/2 + 2 [ 4 2 + 1 2 + 2 2 + 3 2 ] 1/2 = 120 1/2 + 2 1/2 = 4 30 (e) 8. Since k v = [ (–2 k ) 2 + (3 k ) 2 + 0 2 + (6 k ) 2 ] 1/2 = [ 49 k 2 ] 1/2 = 7 | k | , we have | k v | = 5 if and only if k = ±5/7 . 9. (a) (2,5) (–4,3) = (2)(–4) + (5)(3) = 7 (c) (3, 1, 4, –5) (2, 2, –4, –3) = 6 + 2 –16 + 15 = 7 10. (a) Let v = ( x , y ) where v = 1. We are given that v (3, –1) = 0. Thus, 3 x y = 0 or y = 3 x . But v = 1 implies that x 2 + y 2 = x 2 + 9 x 2 = 1 or x = ±1/ 10. Thus, the only possibilities are v = (1/ 10, 3/ 10) or v = (–1/ 10, –3/ 10). You should graph these two vectors and the vector (3, –1) in an xy -coordinate system. (b) Let v = ( x , y , z ) be a vector with norm 1 such that x – 3 y + 5 z = 0 This equation represents a plane through (0,0,0) which is perpendicular to (1, –3, 5). There are infinitely many vectors v which lie in this plane and have norm 1 and initial point (0,0,0).                                                   1 1 3 1 2 2 3 1 2 2 1 2 2 2 2 2 1 2 w w = + + + ( ) = , , , , / , , 1 3 2 2 3 2 3                                   180 Exercise Set 4.1
11. (a) d ( u , v ) = [ (1 – 2) 2 + (–2 – 1) 2 ] 1/2 = 10 (c) d ( u , v ) = [ (0 + 3) 2 + (–2 – 2) 2 + (–1 – 4) 2 + (1 – 4) 2 ] 1/2 = 59 14. (e) Since u v = 0 + 6 + 2 + 0 = 8, the vectors are not orthogonal. 15. (a) We look for values of k such that u v = 2 + 7 + 3 k = 0 Clearly k = –3 is the only possiblity. 16. We must find two vectors x = ( x 1 , x 2 , x 3 , x 4 ) such that x x = 1 and x u = x v = x w = 0. Thus x 1 , x 2 , x 3 , and x 4 must satisfy the equations x 2 1 + x 2 2 + x 2 3 + x 2 4 = 1 2 x 1 + x 2 – 4 x 3 = 0 –x 1 – x 2 + 2 x 3 + 2 x 4 = 0 3 x 1 + 2 x + 5 x 3 + 4 x 4 = 0 The solution to the three linear equations is x 1 = –34 t , x 2 = 44 t , x 3 = –6 t , and x 4 = 11 t . If we substitute these values into the quadratic equation, we get [ (–34) 2 + (44) 2 + (–6) 2 + (11) 2 ] t 2 = 1 or Therefore, the two vectors are ± 1 57 (–34, 44, –6, 11) t = ± = ± 1 3249 1 57                     Exercise Set 4.1 181
17. (a) We have | u v | = |3(4) + 2( 1)| = 10, while u v = [3 2 + 2 2 ] 1/2 [4 2 + (–1) 2 ] 1/2 = 221. (d) Here | u v | = 0 + 2 + 2 + 1 = 5, while u v = [0 2 + (–2) 2 + 2 2 + 1 2 ] 1/2 [(–1) 2 + (–1) 2 + 1 2 + 1 2 ] 1/2 = 6. 18. (a) In this case and 20. By Theorem 4.1.6, we have u v = 1 4 u + v 2 1 4 u v 2 = 1 4 (1) 2 1 4 (5 2 ) = –6. 22. Note that u a = 4 and a 2 = 15. Hence, by Theorem 3.3.3, we have and u u u u a a a = = ( ) proj a 2 2 1 4 1 4 15 4 15 8 15 4 , , , , , , 5 34 15 11 15 52 15 9 5 = , , , proj a 2 u u a a a = = ( ) 4 15 1 1 2 3 , , , A T u . v = = − 9 1 2 6 12 u . v A = = − 3 1 10 18 12 u v . A T = 3 1 2 3 1 4 2 6 = = 3 1 14 26 68 Au . v = = 5 13 2 6 68             182 Exercise Set 4.1
23. We must see if the system 3 + 4 t = s 2 + 6 t = 3 – 3 s 3 + 4 t = 5 – 4 s –1 – 2 t = 4 – 2 s is consistent. Solving the first two equations yield
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