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prelim1_fall05solutions

prelim1_fall05solutions - Math 192 Prelim 1 Solution 1 a In...

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Math 192, Prelim 1. Solution September 29, 2005 1) a) In order for the lines to intersect, there must be a point of intersection at which x 1 = x 2 , so 3 + 2 t = 15 7 s , t = 6 7 s/ 2. The point of intersection also must have y 1 = y 2 , so 3 + 3 t = 2 + s . Substituting t = 6 7 t/ 2 and solving yields s = 2, t = 1. With these values of t and s find z 1 = 1 = z 2 . So there is a point of intersection, and its coordinates are (1 , 0 , 1). b) Let the plane be given by A ( x x o )+ B ( y y o )+ C ( z z 0 ) = 0. The point of intersection is a point in the plane, so can take ( x o , y o , z o ) = (1 , 0 , 1). The direction of L 1 is given by v 1 = 2 i + 3 j + 2 k , that of L 2 by v 2 = 7 i + j + 3 k . The normal to the plane is v 1 × v 2 = i j k 2 3 2 7 1 3 = 7 i 20 j + 23 k . So A = 7, B = 20 and C = 23. The equation for the plane is 7( x 1) 20( y 0) + 23( z + 1) = 0, or 7 x 20 y + 23 z = 16 . 2) The velocity is v = a ( t ) dt = 2 tj + tk + C . At time t = 0, v (0) = 2 i = C , so v ( t ) = 2 i +2 tj + tk . The speed is v ( t ) = 4 + 4 t + t 2 = t +2. The length of the trajectory between t = 0 and t = 3 is d = 3 0 v
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