Question

I Help CH2 HW Begin Date: 8/30/2018 12:00:00 AM--Due Date: 12/13/2018 11:59:00 PMIE (13%) Problem 6: Consider the two vectors, 12/14/2018 12:00:00 AM ent Status 25% Part (a) what is the magnitude of the vector A-B? I l a e> 25% Part (b) what is the angle of the vector A-B, measured counterclockwise in degrees from the positive x-axis? 0 84.1 Correct! Completed Completed Partial 25% Part (c) What is the nitude of the vector 3A +B? Grade Summary cos) cotan0 asin0acos0 Attempts remaining: £ (0se per attempt) detailed view END I give up! 0c 33 x Attempts Remain All

Answer 1

Given two vectors A = 13x + 37.5 y B = 6.9 x - 22y A vector - B vector = (13 x^^+ 37.5 y^^) - (6.9 x^^ -22 y^^) = 6.1 x^^+ 59.5 y^^Magnitude Vertical-bar A vector - B vector Vertical-bar = squareroot 6.1^2 + 59.5^2 = 59.81 Therefore Vertical-bar A - B Vertical-bar = 59.81 Direction angle theta = tan^-1 (y/x) = tan^-1 (59.5/6.1) = 84.1 Therefore theta = 84.1 degree 3A + B = 3 (13x + 37.5 y) + (6.9 x - 22y) = 45.9 x^^+ 90.5 y^^Magnitude = Vertical-bar 3A + B Vertical-bar = squareroot 45.9^2 + 90.5^2 = 101.47 Therefore Vertical-bar 3A + B Vertical-bar = 101.47 \r\n Angle theta = tan^-1 (90.5/45.9) theta = 63.1 degree Therefore Angle of vector 3A + B theta = 63.1 degree