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223midterm220119sol - University of Toronto Department of...

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University of Toronto Department of Mathematics MAT223H1F Linear Algebra I Midterm Exam II November 24, 2011 C. Anghel, S. Arkhipov, S. Shahroki-Tehrani, S. Uppal Duration: 1 hour 30 minutes Last Name: Given Name: Student Number: Tutorial Group: No calculators or other aids are allowed. FOR MARKER USE ONLY Question Mark 1 /10 2 /10 3 /10 4 /10 5 /10 6 /5 TOTAL /55 1 of 7
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1. Consider the points P = (1 , 2 , 3) and Q = (4 , 5 , 6) in R 3 . [6] (a) Find an equation for the plane that passes through the origin and contains the points P and Q . Express your answer in the form ax + by + cz = d . If the normal vector to the plane is n = ( a, b, c ), then n · P = 0 and n · Q = 0 which gives the system a + 2 b + 3 c = 0 4 a + 5 b + 6 c = 0 Solving this system gives n span { (1 , - 2 , 1) } . Therefore the equation of the plane is x - 2 y + z = k for some constant k . Since the plane passes through (0 , 0 , 0) we must have k = 0. Scaling by -3, we get that the equation of the plane is x - 2 y + z = 0. [2] (b) Find the orthogonal projection of ( - 1 , 0 , 8) onto the normal vector to the plane in part (a). Using n = (1 , - 2 , 1) for the normal vector of the plane, setting v = ( - 1 , 0 , 8), we use the equation proj n ( v ) = n · v k n k n = (1 , - 2 , 1) · ( - 1 , 0 , 8) k (1 , - 2 , 1) k n = - 1+0+8 1+4+1 n = 7 6 (1 , - 2 , 1) = ( 7 6 , - 7 3 , 7 6 ) [2] (c) Find the distance from the point ( - 1 , 0 , 8) to the plane in part (a). The distance from ( - 1 , 0 , 8) to the plane is the length of the projection onto the normal direction of any vector v which connects a point on the plane to the point ( - 1 , 0 , 8). Since the plane contains the origin, we can choose v = ( - 1 , 0 , 8) and compute the projection as in part b). The distance is therefore k proj n ( v ) k = k 7 6 (1 , - 2 , 1) k = 7 6 k (1 , - 2 , 1) k = 7 6 1 + 4 + 1 = 7 6 6 2 of 7
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2. Consider the vectors v 1 = (1 , 1 , 0), and v 2 = (1 , 1 , 1) in R 3 .
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