midterm1 A - Solutions to Midterm#1 MAT1341A Question 1(1 1...

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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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midterm1 A - Solutions to Midterm#1 MAT1341A Question 1(1 1...

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