Fall 2007 Final

# Fall 2007 Final - MATH 133 1 The plane has vector equation...

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MATH 133 Final Examination December 17, 2007 1. The plane Π has vector equation x 1 x 2 x 3 = - 4 0 0 + s 1 2 0 + t - 3 0 4 . (a) Find an equation ax 1 + bx 2 + cx 3 = d for the plane Π . ANS: We can take a b c = 1 2 0 × - 3 0 4 = 8 - 4 6 so that an equation for Π 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 , 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) .

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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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