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

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Unformatted text preview: 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 , . 379 , . 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 = ( . 507 , . 406 , . 761) Then a pAN = a p a AN = (0 . 664 , . 379 , . 645) ( . 507 , . 406 , . 761) = ( . 550 , . 832 , . 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 , . 627 , . 348) Now 1 2 ( a AM + a AN ) = 1 2 [(0 . 697 , . 627 , . 348) + ( . 507 , . 406 , . 761)] = (0 . 095 , . 516 , . 207) Finally, a bis = (0 . 095 , . 516 , . 207) | (0 . 095 , . 516 , . 207) | = (0 . 168 , . 915 , . 367) 1 1.27. The surfaces r = 2 and 4, = 30 and 50 , and = 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 drdd = 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 dd + 4 2 60 20 r (sin30 + sin50 ) drd + 2 50 30 4 2 rdrd = 12 . 61 c) Find the total length of the twelve edges of the surface: Length = 4 4 2 dr + 2 50 30 (4 + 2) d + 60 20 (4sin50 + 4sin30 + 2sin50 + 2sin30 ) d = 17 . 49 d) Find the length of the longest straight line that lies entirely within the surface: This will be from A ( r = 2 , = 50 , = 20 ) to B ( r = 4 , = 30 , = 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 , . 52 , 1 . 29) to B (1 . 00 , 1 . 73 , 3 . 46). Thus B A = ( . 44 ,...
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This note was uploaded on 03/30/2010 for the course EE 2317 taught by Professor Wilton during the Spring '10 term at University of Houston.

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

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