A vector a has magnitude of 8units and makes an angle of 45'0 with the positive x-axis vector b also has the same magnitude of 8unuits and directede along the negative x-axis find A) the magnitude and direction of A+B B) the magnitude and direction of A-B
angle of 45'0 with the positive x-axis
Above or below?
If above:
x component = 8 cos 45 -8 = 8 (cos 45 - 1)
y component = 8 sin 45
magnitude = 8 sqrt [ cos 45 -1)^2 + sin^2 45 ]
= 8 sqrt [ 0.0858 + 0.5 ] = 6.12
tan a = sin 45 / (cos 45-1) = .707 / (.707-1) = -2.41
a = 180 - 67.5= 112 deg
do the other the same way