An air traffic controller observes two airplanes approaching the airport. The di

Question

An air traffic controller observes two airplanes approaching the airport. The displacement from the control tower to plane 1 is given by the vector A , which has a magnitude of 220km and points in a direction 32 degrees north of west. The displacement from the control tower to plane 2 is given by the vector B, which has a magnitude of 140km and points 65 degrees east of north. Notice that vector D is the displacement from plane 2 to plane 1.

Vector D = Vector A - Vector B

Find the Magnitude of Vector D.

Find the Direction of Vector D. (theta)

Explanation / Answer

Vector A = 220 cos 32 ( -i) + 220 sin 32 ( j)                = -186.57 i + 116.58 j Vector B = 140 sin 65 ( i) + 140 cos 65 ( j)                = 126.88 i + 59.166 j Vector D = A - B                = ( -186.57 - 126.88 ) i + ( 116.58 -59.166) j                = -313.45 i + 57.41 j Magnitude of D = [ (-313.45) 2 + (57.41) 2 ]                          = 318.66 km Let D amkes an angle with west along north direction then tan = 57.41 / 313.45                                                                                                     = 0.1831                                                                                                   = 10.37 o So, angle with east = 180 -                               = 169.62 o

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