\fMCV4UY - Unit 5 - Vectors Lesson 19 - Review Assignment # 19 &. the symmetric equation of the line. d. the scalar equation of the line C. n = 3 - 6+ n1=-3-66 6 t = 3 -n 6 + = -3 - n t = -2+ 3 6 tz -n-3 6 y = s - 126 12 t = 5 - y y = -7-12+ + = - y +5 12 1 = - 7- 4 12 t= - y-7 12 - 4+ 5 - 1 - 3 - - y-+ 6 12 6 12 ( - 71 +3 ) = 6 ( -y +5 ) 12 ( - 20 - 3 ) = 6(-4-7) - 12 X1 + 36 = - 64+30 - 12 x - 36 = - 64- 42 1 2 7 6 - 6 4 -+ 30 - 36 = 0 12 x - 69 - 42+ 36=0 12 20 - 6 4 - 6 = 0 12 2 - 69 -6 =0\fMCV4UY - Unit 5 - Vectors Lesson 19 - Review Assignment # 19 c. u X B u x V = ( ") ( 2) - (4 ) (6) , (4) (- 5 ) - (-4) (2), ( - 4 ) ( 6 ) - ( 4 ) (- 5 )] = [ 32 7 28 , -47 d. lul 48 101 = / ( - 4 ) + ( LD) + ( -4 ) lul = 6.93 = 16+ 16 + 16 3. Determine the value of k so that u = [- 5, k, 3] and D = [1,2, - 7] are orthogonal. [ 3 marks] ( - 5 ) ( 1 ) + ( k g ( 2 ) + (3 ) ( - 7 ) = 0 - 5 + 2 16 - 21 -6 2K = 2613 :. The value of 14 = 13 16 is 13 4. Given the symmetric equations t = -and t = , find the vector equation of the line. [ 2 marks] y = -1+7t at+l = n [ ny ] = 1 6 , - 1] +t [ 2,7 - 1+7t = yMCV4UY - Unit 5 - Vectors Lesson 19 - Review Assignment # 19 5. A traffic light at an intersection is hanging from two wires of equal length making angles of 109 below the horizontal. The traffic light weighs 2500 N. What are tensions in the wires? [ 3 marks] 10' Sin 80 sin 20" = 250 0 250 ON Sin (80) ( 2 50o ) = sin ( 20) (131 ) Sin ( 80 ) ( 250 0 ) = / u] 10 sin(2 0 ) 80 7 19 8 . 4 6 N = 101 / NI 2500 u . . The tensions in the wire's is N 80 7198 . 46 N 6. Determine the work done by the force, F, in the direction of the displacement, s. [ 2 marks] Work done = Fos IF = 350N 10 F = Force Applied to an Object Is1 = 31 - Objects displacement work done = W = /FIBIcOSO = ( 3 50 ) ( 31 ) cos 10 W = 10685. 20 J Therefore , the work done is 10685 .2 . J 7. Given the force, in Newtons, to be [15,12] for an object moving along the vector [7,3] in metres, find the work done. [ 2 marks] 141 JMCV4UY - Unit 5 - Vectors Lesson 19 - Review Assignment # 19 LESSON 19- ASSIGNMENT # 19 - 30 marks Multiple choice questions -circle the best option- 1 mark each 1. Multiplying a vector by a scalar results in: a. a scalar c. a collinear vector b. a perpendicular vector d. a parallel scalar 2. If A is a vector representing 50 km/h northeast, what is = A ? 10 3 456 a. 30 km/h southwest 30 km/h northwest b. 30 km/h northeast d. 30 km/h southeast 41=30 NE 3. Point A = (1, 3, 4) and point B = (-2, 2, 0). Determine AB. a. (3, 1, 4) (- 2-1), (2- 3 ) , 0-4 c. (-1, 5, 4) b. (-3, -1, -4) d. (1, 5, 4) - 3 - 1 - 4 4. A force vector has a magnitude of 14 N and makes an angle of 20 with the x-axis. What is the magnitude of its horizontal component? a. 13.16 N C. 14.00 N CAH b. 4.79 N d. 14.90 N TO A 5. Under which condition is x . y = 0? a. x and y are in the same direction. 14N b. x and y are perpendicular. 20 C . x and y are in opposite directions. d. Cos 20 = 2 Z None of these conditions force x . y to be zero. 14 # 13 . 16 - 2 N- Be sure to include steps and show your work with all of your calculations. 1. Given the points A(3,5) and B( - 3, - 7), find: [ 2 marks each] a. the vector equation of the line. vo = [3, 5 ]. m - ( - 3 - 3 ) , ( - 7-5 ) = 6-12 my ] = [3, 5 7 + + 5 - 6,-12 Vo = [ - 3 , - 7] [n ,y ] = [-3,-7] + + 5-6,-127 b. the parametric equations of the line [ Use both points to write the eqautions] Using M = 3- 6t X = -3-6+ A ( 3 , 5 ) Using y = 5- 12t y= -7- 126 3 ( - 3 , - 7 )