# chap1 - www.usmanahmad.cjb.net CHAPTER 1 1.1 Given the...

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CHAPTER 1 1.1. Given the vectors M = − 10 a x + 4 a y 8 a z and N = 8 a x + 7 a y 2 a z , find: a) a unit vector in the direction of M + 2 N . M + 2 N = 10 a x 4 a y + 8 a z + 16 a x + 14 a y 4 a z = ( 26 , 10 , 4 ) Thus a = ( 26 , 10 , 4 ) | ( 26 , 10 , 4 ) | = ( 0 . 92 , 0 . 36 , 0 . 14 ) b) the magnitude of 5 a x + N 3 M : ( 5 , 0 , 0 ) + ( 8 , 7 , 2 ) ( 30 , 12 , 24 ) = ( 43 , 5 , 22 ) , and | ( 43 , 5 , 22 ) | = 48 . 6 . c) | M || 2 N | ( M + N ) : | ( 10 , 4 , 8 ) || ( 16 , 14 , 4 ) | ( 2 , 11 , 10 ) = ( 13 . 4 )( 21 . 6 )( 2 , 11 , 10 ) = ( 580 . 5 , 3193 , 2902 ) 1.2. Given three points, A( 4 , 3 , 2 ) , B( 2 , 0 , 5 ) , and C( 7 , 2 , 1 ) : a) Specify the vector A extending from the origin to the point A . A = ( 4 , 3 , 2 ) = 4 a x + 3 a y + 2 a z b) Give a unit vector extending from the origin to the midpoint of line AB . The vector from the origin to the midpoint is given by M = ( 1 / 2 )( A + B ) = ( 1 / 2 )( 4 2 , 3 + 0 , 2 + 5 ) = ( 1 , 1 . 5 , 3 . 5 ) The unit vector will be m = ( 1 , 1 . 5 , 3 . 5 ) | ( 1 , 1 . 5 , 3 . 5 ) | = ( 0 . 25 , 0 . 38 , 0 . 89 ) c) Calculate the length of the perimeter of triangle ABC : Begin with AB = ( 6 , 3 , 3 ) , BC = ( 9 , 2 , 4 ) , CA = ( 3 , 5 , 1 ) . Then | AB | + | BC | + | CA | = 7 . 35 + 10 . 05 + 5 . 91 = 23 . 32 1.3. The vector from the origin to the point A is given as ( 6 , 2 , 4 ) , and the unit vector directed from the origin toward point B is ( 2 , 2 , 1 )/ 3. If points A and B are ten units apart, find the coordinates of point B . With A = ( 6 , 2 , 4 ) and B = 1 3 B( 2 , 2 , 1 ) , we use the fact that | B A | = 10, or | ( 6 2 3 B) a x ( 2 2 3 B) a y ( 4 + 1 3 B) a z | = 10 Expanding, obtain 36 8 B + 4 9 B 2 + 4 8 3 B + 4 9 B 2 + 16 + 8 3 B + 1 9 B 2 = 100 or B 2 8 B 44 = 0. Thus B = 8 ± 64 176 2 = 11 . 75 (taking positive option) and so B = 2 3 ( 11 . 75 ) a x 2 3 ( 11 . 75 ) a y + 1 3 ( 11 . 75 ) a z = 7 . 83 a x 7 . 83 a y + 3 . 92 a z 1

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1.4. given points A( 8 , 5 , 4 ) and B( 2 , 3 , 2 ) , find: a) the distance from A to B . | B A | = | ( 10 , 8 , 2 ) | = 12 . 96 b) a unit vector directed from A towards B . This is found through a AB = B A | B A | = ( 0 . 77 , 0 . 62 , 0 . 15 ) c) a unit vector directed from the origin to the midpoint of the line AB . a 0 M = ( A + B )/ 2 | ( A + B )/ 2 | = ( 3 , 1 , 3 ) 19 = ( 0 . 69 , 0 . 23 , 0 . 69 ) d) the coordinates of the point on the line connecting A to B at which the line intersects the plane z = 3. Note that the midpoint, ( 3 , 1 , 3 ) , as determined from part c happens to have z coordinate of 3. This is the point we are looking for. 1.5. A vector field is specified as G = 24 xy a x + 12 (x 2 + 2 ) a y + 18 z 2 a z . Given two points, P ( 1 , 2 , 1 ) and Q( 2 , 1 , 3 ) , find: a) G at P : G ( 1 , 2 , 1 ) = ( 48 , 36 , 18 ) b) a unit vector in the direction of G at Q : G ( 2 , 1 , 3 ) = ( 48 , 72 , 162 ) , so a G = ( 48 , 72 , 162 ) | ( 48 , 72 , 162 ) | = ( 0 . 26 , 0 . 39 , 0 . 88 ) c) a unit vector directed from Q toward P : a QP = P Q | P Q | = ( 3 , 1 , 4 ) 26 = ( 0 . 59 , 0 . 20 , 0 . 78 ) d) the equation of the surface on which | G | = 60: We write 60 = | ( 24 xy, 12 (x 2 + 2 ), 18 z 2 ) | , or 10 = | ( 4 xy, 2 x 2 + 4 , 3 z 2 ) | , so the equation is 100 = 16 x 2 y 2 + 4 x 4 + 16 x 2 + 16 + 9 z 4 2
1.6. For the G field in Problem 1.5, make sketches of G x , G y , G z and | G | along the line y = 1, z = 1, for 0 x 2. We find G (x, 1 , 1 ) = ( 24 x, 12 x 2 + 24 , 18 ) , from which G x = 24 x , G y = 12 x 2 + 24, G z = 18, and | G | = 6 4 x 4 + 32 x 2 + 25. Plots are shown below.

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