6 polnts hrw10 3p017 go defective Three vectors a b and C ea

-/6 polnts hrw10 3p.017 go defective Three vectors a, b, and C each have a magnitude of B0 m and lie in an xy plane. Their directions relative to the Take these angles counterclockwise to the positive direction of the x axis.) Your Tea s relative to the positive direction of the x axis are 20, 190°, and 260, respectively d.8.a (a) What is the magnitude of the vector a + b + o What s the mpotude (b) what is the angle of the vector (b) What is the angle of the vector+b+c7 + b+ e counterclockwise from the +x-axis (c) What is the magnitude of a- (d) what is the angle ofa-b+c? e counterclockwise from the +x-axis (e) What is the magnitude of a fourth vector d such that (0+ b)-(e, d)-0, fourth vected such that (a + b)-C+-o? (f) what is the angle of counterciockwise from the +x-axis Noed Help? Additional Materials Section 3.2

Solution

a)

a = 80*cos(20) i + 80*sin(20) j

b = 80*cos(190) i + 80*sin(190) j

c =  80*cos(280) i + 80*sin(280) j

So,

a + b + c = (80*cos(20) + 80*cos(190) + 80*cos(280)) i +  (80*sin(20) + 80*sin(190) + 80*sin(280)) j

So,

a + b + c = 10.3 i - 65.3 j

So, magnitude = sqrt(10.3^2 + 65.3^2)

= 66.1 m <----------answer

b)

angle = - atan(65.3/10.3) = -81 deg = 279 deg

c)

a - b + c =  (80*cos(20) - 80*cos(190) + 80*cos(280)) i + (80*sin(20) - 80*sin(190) + 80*sin(280))

So,

a - b + c = 167.9 i - 37.5 j

So, magnitude = sqrt(167.9^2 + 37.5^2)

= 172 m

d)

angle = 360 - atan(37.5/167.9) = 347.4 deg

e)

(a+ b) - (c+d) = 0

So, d = a+b-c

So, d =  (80*cos(20 deg) + 80*cos(190 deg) - 80*cos(280 deg)) i + (80*sin(20 deg) - 80*sin(190 deg) - 80*sin(280 deg)) j

So, d = -17.5 i + 120 j

So, magnitude = sqrt(17.5^2 + 120^2)

= 121.3 m

f)

angle = 180 - atan(120/17.5) = 98.3 deg