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MVC SHOW ALL WORK !! No Calculator Quiz #1 NAME: #1(3 pt's) Find the parametric equations of the line that contains the point (1,2,3) and is parallel to x 2 z+1 , y 4. the line with symmetric equations 5 3 x 1 5t M (5,0,3) is parallel to the line. So the parametric equations are: y 2 z 3 3t #2(4 pt’s). Let x = (3, 4, –1) and y = (–5, 2, 1). a. Find x 32 42 (1)2 26 . b. 8 xy 12 16 4 (3, 4, 1) , , Find projx y x 26 xx 13 13 13 x 1 2sin t #3(4 pts) Describe the curve given parametrically by y 3 sin t , 0 t 2 . (You need not z 2 3sin t graph the curve). Be as explicit as possible. Describe the shape of the curve and how it is traced out as t goes from 0 to 2 . This is a line segment. Since 1 sin t 1, this will be the line segment from ( x, y, z ) (1 2, 3 1, 2 3) (1, 4,5) to the point (3, 2, 1) . It is traced out from (1, 3, 2) to (3, 2,1) then back through (1, 3, 2) to (1, 4,5) and finally back to the point (1, 3, 2) MVC #4(4 pt’s) Let u , v , w 3 . Determine whether each statement is true or false. If true, prove it. If false, give an example that demonstrates it is false. i. If u 0 and u w u v , then w v False: Let u (1,0,0), w (3,1,1)and v (3, 211,190). Then u w 3 u v , but w v . ii. u v u v False: If u iˆ and v ˆj, then u v 0 and u v 1. Note that this is true when the vectors are parallel. #5(5 pt's) Let x (5, a,3a) and y (10, 4, 6a) a. Find all values of a such that x and y orthogonal. We know that if x and y orthogonal, then x y 50 4a 18a 2 0 9a 2 2a 25 0 no such a exists. b. Find all values of a such that x and y parallel. If x and y parallel, then y is a scalar multiple of x. Looking at the first component, y 2x . Then, 2a 4 and 2(3a) 6a. Therefore, a 2 . MVC #6 (3 pt’s) Prove that the midpoints of a quadrilateral in 3 form a parallelogram. Proof. Let ABCD be the quadrilateral as shown. (Note that is not necessary that the four vertices lie in the same plane). Let E , F , G and H be the midpoints of the sides ad pictured. A E B H F D G C To show that quadrilateral EFGH is a parallelogram, it suffices to show that EF HG and FG EH . 1 1 AB BC = AD DC and it follows that We note that: AB BC = AD DC . Then, 2 2 EB BF = HD DG . Observing the figure, this gives us immediately that EF HG . The proof that FG EH is similar starting with BC CD = BA AD MVC