Fall 2007 Final

Fall 2007 Final - MATH 133 Final Examination December 17,...

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Unformatted text preview: MATH 133 Final Examination December 17, 2007 1. The plane has vector equation x 1 x 2 x 3 = - 4 + s 1 2 + t - 3 4 . (a) Find an equation ax 1 + bx 2 + cx 3 = d for the plane . ANS: We can take a b c = 1 2 - 3 4 = 8- 4 6 so that an equation for is 8 x 1- 4 x 2 + 6 x 3 =- 32 or 4 x 1- 2 x 2 + 3 x 3 =- 16 . An alternate way to do the question is to solve the equations s- 3 t = x 1 + 4 , 2 s = x 2 for s,t to get s = x 2 / 2 ,t = x 2 / 6- x 1 / 3- 4 / 3 and use the fact that x 3 = 4 t to get x 3 = x 2 / 2- 4 x 1 / 3- 16 / 3 or 4 x 1- 2 x 2 + 3 x 3 =- 16 . (b) Find the point Q in the plane 2 x + 3 y + z = 10 which is closest to the point P (7 , 7 , 3) . ANS: The line perpendicular to the given plane and passing through the point (7 , 7 , 3) has the equation x = 7 + 2 t,y = 7 + 3 t,z = 3 + t . The line meets the plane in the point corresponding to the parameter value t satisfying 2(7 + 2 t ) + 3(7 + 3 t ) + (3 + t ) = 10 . This gives t =- 2 which makes the closest point (3 , 1 , 1) . An alternate way to do the question is to use the fact that Q (5 , , 0) is on the plane and to note that the closest point can be obtained by adding to the coordinates of P the projection of- PQ = (- 2 ,- 7 ,- 3) onto the normal (2 , 3 , 1) of the given plane. This gives the point (7 , 7 , 3) +- 28 14 (2 , 3 , 1) = (3 , 1 , 1) . MATH 133 Final Examination December 17, 2007 2. (a) Find the equation of the line passing through the points A (1 , 2 , 3) and B (2 , 1 , 5) . ANS: A direction vector for the line is- AB = (1 ,- 1 , 2) so that the equation for the line in parametric form is x = 1 + t,y = 2- t,z = 3 + 2 t . (b) Find the distance between the line in part (a) and the line x = 2- 2 t,y = 4 + 2 t,z = 7- 4 t . ANS: The point C (2 , 4 , 7) lies on the given line. If A and B are the points in part (a) then the distance d between the two lines is area of the parallelogram with sides parallel to- AB and- AC divided by the length of- AB . Now- AB - AC = (1 ,- 1 , 2) (1 , 2 , 4) = (- 8 ,- 2 , 3) and the area of the parallelogram is ||- AB - AC || = 77 so that d = 77 / 6 = p 77 / 6 . Alternatively, d is the length of AC-- AC - AB- AB - AB- AB = (1 , 2 , 4)- 7 6 (1 ,- 1 , 2) = 1 6 (- 1 , 19 , 10) which is 462 / 6 = p 77 / 6 . MATH 133 Final Examination December 17, 2007 3. Let A be the matrix A = 0 0 1 1 1 0 0 4 4 4 1 2- 3- 8 0 1 2- 1- 6 2 ....
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This note was uploaded on 12/01/2010 for the course MATH 133 taught by Professor Klemes during the Fall '08 term at McGill.

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Fall 2007 Final - MATH 133 Final Examination December 17,...

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