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sm - 1 Problem Solutions for E-text 1.3 The vector from the...

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Problem Solutions for E-text 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 1.17. Point A ( 4 , 2 , 5) and the two vectors, R AM = (20 , 18 , 10) and R AN = ( 10 , 8 , 15), define a triangle. a) Find a unit vector perpendicular to the triangle: Use a p = R AM × R AN | R AM × R AN | = (350 , 200 , 340) 527 . 35 = (0 . 664 , 0 . 379 , 0 . 645) The vector in the opposite direction to this one is also a valid answer. b) Find a unit vector in the plane of the triangle and perpendicular to R AN : a AN = ( 10 , 8 , 15) 389 = ( 0 . 507 , 0 . 406 , 0 . 761) Then a pAN = a p × a AN = (0 . 664 , 0 . 379 , 0 . 645) × ( 0 . 507 , 0 . 406 , 0 . 761) = ( 0 . 550 , 0 . 832 , 0 . 077) The vector in the opposite direction to this one is also a valid answer. c) Find a unit vector in the plane of the triangle that bisects the interior angle at A : A non-unit vector in the required direction is (1 / 2)( a AM + a AN ), where a AM = (20 , 18 , 10) | (20 , 18 , 10) | = (0 . 697 , 0 . 627 , 0 . 348) Now 1 2 ( a AM + a AN ) = 1 2 [(0 . 697 , 0 . 627 , 0 . 348) + ( 0 . 507 , 0 . 406 , 0 . 761)] = (0 . 095 , 0 . 516 , 0 . 207) Finally, a bis = (0 . 095 , 0 . 516 , 0 . 207) | (0 . 095 , 0 . 516 , 0 . 207) | = (0 . 168 , 0 . 915 , 0 . 367) 1

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1.27. The surfaces r = 2 and 4, θ = 30 and 50 , and φ = 20 and 60 identify a closed surface. a) Find the enclosed volume: This will be Vol = 60 20 50 30 4 2 r 2 sin θdrdθdφ = 2 . 91 where degrees have been converted to radians. b) Find the total area of the enclosing surface: Area = 60 20 50 30 (4 2 + 2 2 )sin θdθdφ + 4 2 60 20 r (sin30 + sin50 ) drdφ + 2 50 30 4 2 rdrdθ = 12 . 61 c) Find the total length of the twelve edges of the surface: Length = 4 4 2 dr + 2 50 30 (4 + 2) + 60 20 (4sin50 + 4sin30 + 2sin50 + 2sin30 ) = 17 . 49 d) Find the length of the longest straight line that lies entirely within the surface: This will be from A ( r = 2 , θ = 50 , φ = 20 ) to B ( r = 4 , θ = 30 , φ = 60 ) or A ( x = 2sin50 cos20 , y = 2sin50 sin20 , z = 2cos50 ) to B ( x = 4sin30 cos60 , y = 4sin30 sin60 , z = 4cos30 ) or finally A (1 . 44 , 0 . 52 , 1 . 29) to B (1 . 00 , 1 . 73 , 3 . 46). Thus B A = ( 0 . 44 , 1 . 21 , 2 . 18) and Length = | B A | = 2 . 53 2.5. Let a point charge Q 1 25 nC be located at P 1 (4 , 2 , 7) and a charge Q 2 = 60 nC be at P 2 ( 3 , 4 , 2). a) If = 0 , find E at P 3 (1 , 2 , 3): This field will be E = 10 9 4 π 0 25 R 13 | R 13 | 3 + 60 R 23 | R 23 | 3 where R 13 = 3 a x +4 a y 4 a z and R 23 = 4 a x 2 a y +5 a z . Also, | R 13 | = 41 and | R 23 | = 45. So E = 10 9 4 π 0 25 × ( 3 a x + 4 a y 4 a z ) (41) 1 . 5 + 60 × (4 a x 2 a y + 5 a z ) (45) 1 . 5 = 4 . 58 a x 0 . 15 a y + 5 . 51 a z b) At what point on the y axis is E x = 0? P 3 is now at (0 , y, 0), so R 13 = 4 a x + ( y + 2) a y 7 a z and R 23 = 3 a x + ( y 4) a y + 2 a z . Also, | R 13 | = 65 + ( y + 2) 2 and | R 23 | = 13 + ( y 4) 2 .
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sm - 1 Problem Solutions for E-text 1.3 The vector from the...

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