Question

Three vectors a → , b → , and c → , each have a magnitude of 48.0 m and lie in an xy plane. Their directions relative to the positive direction of the x axis are 32.0 ˚, 193 ˚, and 313 ˚, respectively. What are (a) the magnitude and (b) the angle of the vector a → + b → + c → (relative to the +x direction in the range of (-180°, 180°)), and (c) the magnitude and (d) the angle of a → - b → + c → in the range of (-180°, 180°)? What are (e) the magnitude and (f) the angle (in the range of (-180°, 180°)) of a fourth vector d → such that ( a → + b → ) - ( c → + d → ) = 0 ?

Answer 1

a)

= (48 Cos32) i + (48 Sin32) j = 40.71 i + 25.44 j

= (48 Cos193) i + (48 Sin193) j = - 46.77 i - 10.8 j

= (48 Cos313) i + (48 Sin313) j = 32.74 i - 35.1 j

Sum of the vectors is given as

= + +

= (40.71 i + 25.44 j) + (- 46.77 i - 10.8 j) + (32.74 i - 35.1 j )

= 26.68 i - 20.46 j

|| = magnitude = sqrt((26.68)² + (- 20.46)²) = 33.62

b)

Direction :

= - tan^-1(20.46/26.68) = - 37.5 deg

c)

= - +

= (40.71 i + 25.44 j) - (- 46.77 i - 10.8 j) + (32.74 i - 35.1 j )

= 120.22 i + 1.14 j

|| = magnitude = sqrt((120.22)² + (1.14)²) = 120.225

d)

Direction :

= tan^-1(1.14/120.22) = 0.54 deg

e)

+ - + = 0

(40.71 i + 25.44 j) + (- 46.77 i - 10.8 j) - (32.74 i - 35.1 j ) + = 0

= - (40.71 i + 25.44 j) - (- 46.77 i - 10.8 j) + (32.74 i - 35.1 j )

= 38.8 i - 49.74 j

|| = magnitude = sqrt((38.8)² + (- 49.74)²) = 63.08

f)

Direction :

= tan^-1(- 49.74/38.8) = - 52 deg