Question

displacement vectors shown in the figure: Vector A has a magnitude of 8.30 km and a direction that makes an angle θ-230° to the s, and vector ft of the positive y-axis, vector B has a magnitude of 6.00 km and a direction that makes an angle of a -340 above the positive x-axi as a magnitude of 4.20 km and a direction that makes an angle B -63.0° below the negative x axis, Determine the magnitude of km

Answer 1

Vertical-bar A^vector Vertical-bar = 8.30 km, theta = 23.0 degree Vertical-bar B^vector Vertical-bar = 6.00 km, alpha = 34.0 degree Vertical-bar C^vector Vertical-bar = 4.20 km, beta = 63.0 degree A^vector = (8.30 km) (-sin 23 degree i^^+ cos 23 degree j^^) B^vector = (6.00 km) (cos 34 degree i^^+ sin 34 degree j^^) C^vector = (4.20 km) (-cos 63 degree i^^- sin 63 j^^) therefore D^vector = A^vector - B^vector + C^vector = [(-8.3 sin 23 degree - 6 cos 34 degree - 4.2 cos 63 degree) i^^+ (8.3 cos 23 degree - 6 sin 34 degree - 4.2 sin 34 degree 63 degree) j^^] therefore D6vector = A6vector - B^vector + C^vector = (-10.1 i^^+ 0.543 j^^) km Vertical-bar D^vector Vertical-bar = squareroot (10.1)^2 + (0.543)^2 km = 10.13 km therefore Magnitude D^vector = A^vector + C^vector = 10.13 km 10.13 km