# ism4 - Chapter 4 Vectors in Rn Section 4.1 p 227 2 y 4 u1...

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Chapter 4 Vectors in R n Section 4.1, p. 227 2. x y O 3 4 u 1 u 2 u 3 u 4 4. ( 1 , 3) x y O (1 , 2) ( 1 , 3) 6. (a) u + v =(1 , 7); u v =( 3 , 1); 2 u 2 , 6); 3 u 2 v 7 , 1). (b) u + v , 1); u v 9 , 5); 2 u 8 , 6); 3 u 2 v 22 , 13). (c) u + v , 2); u v =(5 , 2); 2 u =(6 , 4); 3 u 2 v = (13 , 6). 8. (a) x = 2, y = 9. (b) x = 6, y = 8. (c) x =5, y = 25 2 . 10. (a) 13. (b) 3. (c) 41. (d) 13. 12. (a) 3. (b) 20. (c) 18. (d) 5. 14. Impossible 16. 6. 18. 41 2 . 20. (a) ³ 1 5 , 2 5 ´ . (b) (0 , 1). (c) ³ 1 10 , 3 10 ´ . 22. (a) 0. (b) 1 2 41 . (c) 4 5 13 . (d) 1 2 . 24. (a) u 1 and u 4 , u 1 and u 6 , u 3 and u 4 , u 3 and u 6 , u 4 and u 5 , u 5 and u 6 . (b) u 1 and u 5 , u 4 and u 6 . (c) u 1 and u 3 , u 3 and u 5 .

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68 Chapter 4 26. a = ± 2. 28. (a) ± 3 2 ² . (b) ± 2 0 ² . (c) ± 2 3 ² . 30. x y W R 100 km/hr Plane heading 260 km/hr Resultant speed: 240 km/hr. T.1. Locate the point A on the x -axis which is x units from the origin. Construct a perpendicular to the x -axis through A . Locate B on the y -axis y units from the origin. Construct a perpendicular through B . The intersection of those two perpendiculars is the desired point in the plane. T.2. ( x, y )+(0 , 0) = ( x +0 ,y +0)=( x, y ). T.3. ( x, y )+( 1)( x, y )=( x, y x, y x x, y y )=(0 , 0). T.4. ² c u ² = p ( cx ) 2 +( cy ) 2 = c 2 p x 2 + y 2 = | c u ² . T.5. ² u ² = ³ ³ ³ ³ 1 ² x ² x ³ ³ ³ ³ = 1 ² x ² ² x ² =1. T.6. (a) 1 u =1 · ( x, y )=(1 · x, 1 · y x, y )= u . (b) ( rs ) u =( )( x, y rsx, rsy r ( sx, sy r ( s u ). T.7. (a) u · u = ² u ² 2 = x 2 + y 2 0; u · u = 0 if and only if x = 0 and y = 0, that is, u = 0 . (b) ( x 1 1 ) · ( x 2 2 x 1 x 2 + y 1 y 2 = x 2 x 1 + y 2 y 1 x 2 2 ) · ( x 1 1 ). (c) [( x 1 1 x 2 2 )] · ( x 3 3 x 1 + x 2 ) x 3 y 1 + y 2 ) y 3 = x 1 x 3 + y 1 y 3 + x 2 x 3 + y 2 y 3 = ( x 1 1 ) · ( x 3 3 x 2 2 ) · ( x 3 3 ). (d) ( cx 1 ,cy 1 ) · ( x 2 2 cx 1 x 2 + cy 1 y 2 x 1 1 ) · ( cx 2 2 c ( x 1 x 2 + y 1 y 2 c [( x 1 1 ) · ( x 2 2 )]. T.8. If w · u =0= w · v , then w · ( r u + s v r ( w · u )+ s ( w · v )=0+0=0. T.9. If u and v are parallel, then there exists a nonzero scalar k such that v = k u .Thu s cos θ = u · v ² u ²² v ² = u · ( k u ) ² u k u ² = k ( u · u ) ² u ² p ( k u ) · ( k u ) = k ² u ² 2 ² u ² k 2 ² u ² = k k 2 = k ± k = ± 1 .
Section 4.2 69 Section 4.2, p. 244 2. (a) u + v = 5 2 3 , u v = 1 2 5 ,2 u = 4 0 8 ,3 u 2 v = 0 4 14 . (b) u + v = 1 6 2 2 , u v = 5 4 8 2 u = 6 10 6 0 u 2 v = 13 13 19 4 . 4. (a) a = 12, b = 9. (b) a =2, b = 1, c d = 4. (c) a =7, b =4. 6. z x y (3 , 1 , 2) (1 , 0 , 2) (1 , 0 , 0) (0 , 0 , 4) (0 , 2 , 0) 8. (a) 2 3 3 . (b) 1 0 1 . (c) 4 6 8 . (d) 1 1 2 . 10. (a) 14. (b) 30. (c) 10. (d) 5. 12. (a) 5. (b) 6. (c) 13. (d) 30. 14. Impossible. 16. a = ± 11. 18. (a) u · u =1 2 +2 2 +3 2 1=1+4+9=14 0.

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## This note was uploaded on 06/25/2008 for the course MATH 22a taught by Professor Chuchel during the Spring '08 term at UC Davis.

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ism4 - Chapter 4 Vectors in Rn Section 4.1 p 227 2 y 4 u1...

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