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m24w10midtermsolns

# m24w10midtermsolns - Math 24 Winter 2010 Sample Solutions...

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Math 24 Winter 2010 Sample Solutions to the Midterm (1.) (a.) Find a basis { v 1 , v 2 } for the plane P in R 3 with equation 3 x + 2 y - z = 0. We can take any two non-collinear vectors in the plane, for instance v 1 = (1 , 0 , 3) and v 2 = (0 , 1 , 2). (b.) You know from multivariable calculus that the vector v 3 = (3 , 2 , - 1) is perpendicular to the plane P . Therefore β = { v 1 , v 2 , v 3 } is linearly independent, and forms an ordered basis for R 3 . Let T : R 3 R 3 be the perpendicular projection onto the plane P . In other words, T ( v ) is the perpendicular projection of v onto P . What is [ T ] β ? As v 1 and v 2 are in the plane, T ( v 1 ) = v 1 and T ( v 2 ) = v 2 . As v 3 is a vector perpendicular to the plane, its projection is the origin, T ( v 3 ) = 0. Therefore [ T ( v 1 )] β = [ v 1 ] β = 1 0 0 , [ T ( v 2 )] β = [ v 2 ] β = 0 1 0 , [ T ( v 3 )] β = [0] β = 0 0 0 , [ T ] β = 1 0 0 0 1 0 0 0 0 . (c.) Let α be the standard ordered basis for R 3 . Find the change of coordinate matrices Q β α that changes α coordinates into β coordinates, and Q α β that changes β coordinates into α coordinates. Do not use matrix inversion (if you know how to invert matrices) to do this problem. Find each matrix by explicitly computing the coordinates of the appropriate vectors in the appropriate bases. If you wish, you can check your work by verifying that Q β α Q α β = I . Q α β = 1 0 3 0 1 2 3 2 - 1 , the matrix whose columns are the α (standard) coordinates of the vectors in β . To find the β coordinates of the vectors in α (the standard basis vectors) we need to solve the vector equations: (1 , 0 , 0) = a (1 , 0 , 3) + b (0 , 1 , 2) + c (3 , 2 , - 1) (0 , 1 , 0) = a (1 , 0 , 3) + b (0 , 1 , 2) + c (3 , 2 , - 1) (0 , 0 , 1) = a (1 , 0 , 3) + b (0 , 1 , 2) + c (3 , 2 , - 1) When we do this, we get (1 , 0 , 0) = 5 14 (1 , 0 , 3) - 6 14 (0 , 1 , 2) + 3 14 (3 , 2 , - 1) (0 , 1 , 0) = - 6 14 (1 , 0 , 3) + 10 14 (0 , 1 , 2) + 2 14 (3 , 2 , - 1) 1

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(0 , 0 , 1) = 3 14 (1 , 0 , 3) + 2 14 (0 , 1 , 2) - 1 14 (3 , 2 , - 1), that is, [(1 , 0 , 0) β = 5 14 - 6 14 3 14 [(0 , 1 , 0)] β = - 6 14 10 14 2 14 [(0 , 0 , 1)] β = 3 14 2 14 - 1 14 . Now, using these coordinates as the columns of Q β α , we have Q β α = 5 14 - 6 14 3 14 - 6 14 10 14 2 14 3 14 2 14 - 1 14 . (d.) Find the matrix of T in the standard basis, [ T ] α . If you want to use the result of part (b) but were not able to do part (b), you may pretend [ T ] β = 1 0 0 0 1 0 0 0 2 . This is not the correct answer to part (b). [ T ] α = [ T ] α α = Q α β [ T ] β β Q β α = 1 0 3 0 1 2 3 2 - 1 1 0 0 0 1 0 0 0 0 5 14 - 6 14 3 14 - 6 14 10 14 2 14 3 14 2 14 - 1 14 . [ T ] α = 5 14 - 6 14 3 14 - 6 14 10 14 2 14 3 14 2 14 13 14 . (e.) Use your answers to find the perpendicular projection of (2 , 1 , 1) onto the plane P . If you want to check your work here, note that this point should in fact be on the plane P , and the line between it and (2 , 1 , 1) should be perpendicular to P . If you used the wrong answer to part (b) supplied in part (d), the point you found will not be on P , but the line between it and (2 , 1 , 1) will still be perpendicular to P .
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m24w10midtermsolns - Math 24 Winter 2010 Sample Solutions...

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