1)Find the magnitude of the vector difference A-B, 2)Find the direction of the v

Question

1)Find the magnitude of the vector difference A-B,

2)Find the direction of the vector difference A-B,

3)Find the magnitude of the vector difference B-A.

Please someone help, WILL RATE 5 stars. Thanks

Explanation / Answer

In order to answer these questions, you first need to construct vectors A and B in vector notation. First notice that vector A is entirely in the negative y direction, so it would be A = - (8m)y-hat Vector B is made at a 30 degree angle to the positive y axis, so its y component would be By = (15m)cos(30) = 13m and its x component would be (15m)sin(30) = 7.5m So, B is B = (7.5m)x-hat + (13m)y-hat 1) The vector difference is A - B = -(8m)y-hat - (7.5m)x-hat - (13m)y-hat = - (7.5m)x-hat - (21m)y-hat So, the magnitude is |A - B| = sqrt( (-7.5m)^2 + (-21m)^2) = 22.3m 2) First note that vector A - B is in the third quadrant (negative x and negative y), so if I use the tangent method, then it will give me an angle below the negative x-axis. With that in mind, tan(theta) = (21m)/(7.5m) --> theta = 70.35 degrees below the negative x-axis This is equivalent to 180 degrees + 70.35 degrees = 250.35 degrees counter clockwise from the positive x-axis. 3) B - A = (7.5m)x-hat + (13m)y-hat - (-8m)y-hat = +(7.5m)x-hat + (21m)y-hat This is just - (A - B), as one would expect, so the magnitude is the same: |B - A| = |A - B| = 22.3m Hope this helps.

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