Question

I keep getting different advice on how to tackle this question. Can anyone explain how to do this?

Given two vectors A and B, with magnitudes |A| = 80 and |B| = 40 and directions (from the
x-axis) θA=60° and θB=135°, find the magnitude of (A-B).

Answer 1

Hey there,

This is a vector addition problem. \[\overrightarrow A + (-\overrightarrow B)\]
First we need to break down each vector into Cartesian coordinates (x and y components) using trigonometric functions sine and cosine:
\[V_x = |V|\cos{V_\theta}\]\[V_y = |V|\sin{V_\theta}\] So.. \[A_x = 80\cos{60^o} = 40.00\]\[A_y = 80\sin{60^o} = 69.28\]\[B_x = 40\cos{135^o} = -28.28\]\[B_y = 40\sin{135^o} = 28.28\]
Now, we need to add the components together as dictated by the original expression:
\[V_c = A_c + (-B_c)\] So.. \[V_x = A_x + (-B_x) = 40.00 - (-28.28) = 68.28\]\[V_y = A_y + (-B_y) = 69.28 - 28.28 = 41.00\]

Now that we have the components of our new vector, we need to resolve it's magnitude and direction.

It's magnitude can be found using the Pythagorean theorem: \[|V| = \sqrt{V_x^2 + V_y^2} = \sqrt{68.28^2 + 41.00^2} = 79.13\] And it's direction can be found using arc tangent: \[V_d = \arctan{\frac{V_y}{V_x}} = \arctan{\frac{41.00}{68.28}} = 30.98^o\]

Answer 2

id just apply the law of cosines, which is a more generalization of the Pythag thrm.
\[|C|^2=|A|^2+|B|^2-2|A||B|~cos~\alpha\]
\[|C|^2=80^2+40^2-2(80)(40)~cos(135^o-60^o)\]
\[|C|^2=8000 -6400~cos(75^o)\]
\[|C|=10\sqrt{80 -64~cos(75^o)}\]