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Unformatted text preview: Solutions to Midterm #1, MAT1341A Question # 1: (1+1+1+1+1+1=6 points) Write the answer for each of the following questions in the box provided. (a) Simplify: (2 i )(2 + 3 i ). Solution: (2 i )(2 + 3 i ) = 4 3 i 2 + (6 2) i = 7 + 4 i (b) Compute the absolute value  1 5 i  . Solution:  1 5 i  = √ 1 + 25 = √ 26 (c) Simplify 1 1 5 i . Solution: 1 1 5 i = 1 + 5 i  1 5 i  2 = 1 + 5 i 26 = 1 26 + 5 26 i (d) Calculate the dot product of the following vectors: (2 , 3 , 1 , 1) · (5 , 2 , 5 , 1) Solution: (2 , 3 , 1 , 1) · (5 , 2 , 5 , 1) = 10 6 5 + 1 = 0 (e) Are the two vectors in part (d) (above) orthogonal? (Yes or no) Solution: Their dot product is zero so YES they are orthogonal. (f) Calculate the cross product (2 , 1 , 4) × (6 , 2 , 1). Solution: (2 , 1 , 4) × (6 , 2 , 1) = (( 1 ( 8)) , (2 24) , 4 ( 6)) = (7 , 22 , 2) Question # 2: (4 points) Find the point of intersection of the following two lines in R 4 : L 1 = 1 2 2 4 + t  1 3 5  t ∈ R and L 2 = 3 20 10 6 + s  1 3 2  s ∈ R . Write your answer in the box below. 1 2 Solution: We are looking for a common point on the two lines. The points on L 1 are parametrized by t , those on line L 2 are parametrized by s , so we are looking for values s, t which gives us an equality: 1 2 2 4 + t  1 3 5 = 3 20 10 6 + s  1 3 2 . This is equivalent to solving the equations 1 t = 3 s, 2 + 3 t = 20 3 s, 2 = 10 + 2 s, 4 5 t = 6 . The third implies s = 4, while the fourth implies t = 2. Now t = 2 corresponds to the point ( 1 , 8 , 2 , 6) on L 1 and s = 4 corresponds to the point ( 1 , 8 , 2 , 6) on L 2 , so indeed, the lines intersection and the point of intersection is ( 1 , 8 , 2 , 6) . Question # 3: (3+3=6 points) Let V = a b c  a, b, c ∈ R , with addition and scalar multiplication defined by the following: a b c ⊕ d e f = a + d be cf t ?...
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 Winter '10
 DAIGLE
 Vector Space, #, M2,2

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