6. Determine the vertical and horizontal components of each vector. [6 marks] a. 810 N at an incline of 15 from horizontal b. 80 m/s, 60 clockwise from vertical 7. Given a = 2, -10 and b = [-8, 23 , find: a. 3a - 2b [2 marks] b. a . b [1 marks] c. a + b [2 marks] d. proj a [3 marks] b e. The angle between a and 6 [2 marks]17. Given the equations x, y, z = 7, -15, -4 + 8 1, -2,4 and a, y, 2 = [-15, -26, 23] + t[5, 1, -3], a. Identify whether the equations represent lines or planes. How do you know? [2 marks] b. Solve the system of equations to find the point of intersection. [5 marks]17. a) b) The equation represents a line is m= = a= + x b= is on an equation of a line where a = [x,y.z] is a point on line and b = = [ p, g, m] is parallel reactor relative to the line of and x = [x, y, Z] is general point reactor. For the point of intersection x.y,z] = [7, -15, -4] + s[1,-2,4] =[-15, -26,23] + 5 [5,1,-3] [x.y.z] = [7+s, -15-23, -4 + 4s] =[-15+ 5t -26+ 1, 23 - 31] Therefore, 7T+s5=-15 +5t bt-s =221 -15-25 =-26 + =2s+t=11-2 Point of intersections are -1 and -2 8 x= 1+t Y =4 -t Z=-2 +3t Sub the values in the plane equation 2x - 5y + 8t - 28 =0 2(1+ t) - 5(4 - t) + 8 (-2 +3t) -28 = 0 31 t -62 =0 T =2 Finding the equation coordinates (x,y,z) X = 1 +t = 1 + 2 =3 Y =4 - t =4 - 2 =2 Z = - 2+6 =4 The intersection at the line and the plane is at (3,2,4)18. Find the intersection of [z, y, z| = [1,4,2] + [1,1, 3] and 2z 5y + 8z 28 = 0. [4 marks] 19. Find the distance between the point (10, 100, 1000) and the plane [z,y,2] = [1,1,1] + s[5, 5,3] + t[1,1, 3] [5 marks] 19 D = laxo + byot cro +DI VA +bc Points to consider in the problem p(x0,y0,zo) And the plane equation Ax +By + Cz +d = 0 Given p(10, 100, 1000) [x, y,z] = [1,1, 1] + s[5,-5,3] + t[-1,2,3] A= 5, B =-5, c=3 The values of D at the point of [1, 1,1] D= -Ax -By - Cz D= -(5x1) - (-5 x1) - (3x1) D= -5+5-3 D= -3 The equation for the plane = 5x -5y + 3z -3 = 0 D = (6x10) -5(100) +13 x1000)| 25 + 25 + 9 D = 125471 2547 = 331.59 59 159 The distance between the given point and the plane is 331.59 units.6. a) Given: 810 N at an incline of 15 from horizontal Fn is the horizontal component and the fr is the vertical component. Fn = fx cos(15 degrees) Fr = fx sin(15 degrees) Fn = 810 N x cos (15 degrees) = 782.399 N Fr = 810 N x sin (15 degrees) = 209.643 N b) 80 m/s, 60 clockwise from vertical Vv = verticalcomponent Vh = horizontal component 0 = 360 - 60 degrees = 300 degrees Vh = V x cos e Vv = V x sin e Vh = 80 x cos (300) = 40 m/'s Vv = 80 x sin (300) = -69.2 m/s 7. a) (3a=) - (2b=) = 3 (2, -10) - 2 (-8, 23) = [ 6, -30] - [-16, 46] = [6 + 16, -30 - 46] = [22. -761\f