Vector Subtract Given Magnitude

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To find the magnitude of vector C, defined as C = A - B, where A has a magnitude of 40 and B has a magnitude of 50 at an angle of 60 degrees, the law of cosines must be applied. The initial assumption that the magnitude of C is simply 10 is incorrect, as vector magnitudes do not subtract directly unless the vectors are parallel. By decomposing vector A into its horizontal and vertical components using trigonometric functions, the components are calculated as 20 and 34.64, respectively. After adjusting for the negative component of vector B, the resultant components are combined, leading to a final magnitude of 45.83 for vector C. This demonstrates the importance of vector decomposition and the law of cosines in vector subtraction problems.
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Homework Statement


Let's say i was given Vector A and B. The Angle between them is 60 degrees. Vector A's magnitude is 40 and Vector B's magnitude is 50. Find magnitude of vector C, if C = vector A - vector B.

Homework Equations


I'm given the magnitudes, and need to find magnitude of C

The Attempt at a Solution


I'm kinda confused cause I don't think this is a simple problem. But, I thought that magnitude of C would be 10. Since A-B = -10 , and 10 is the magnitude[/B]
 
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Can you draw a diagram for the situation?
 
upload_2017-12-17_23-38-33.png
 

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By which side of the triangle the vector A-B is represented? What do you get if you properly use law of cosines?
 
Magnitudes don't add that way unless the vectors are parallel and so you need to solve the triangle using the law of cosines.

For example, if the angle between A and B were 0, then by the law of cosines

c^2=a^2+b^2-2ab\cos{(0)}= (a-b)^2 <br /> \\<br /> c = |a-b|<br /> <br />

Which is what you tried to do but the angle is not zero in this case which is why it's wrong
 
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RedDelicious said:
Magnitudes don't add that way unless the vectors are parallel and so you need to solve the triangle using the law of cosines.

For example, if the angle between A and B were 0, then by the law of cosines

c^2=a^2+b^2-2ab\cos{(0)}= (a-b)^2<br /> \\<br /> c = |a-b|<br /> <br />

Which is what you tried to do but the angle is not zero in this case which is why it's wrong
Thanks. I definitely did not learn this from my physics class.
 
I thought of this problem as C = A + (-B). This means that it is the same as adding a vector, just the vector you are adding is negative. In order to add the vectors, we must split them into their components. We know that vector A is at an angle of 60 and therefore has X and Y components, while B is parallel to the horizontal and only has an X component equal to its magnitude.

To split vector A, we use Cos and Sin to find the vertical and horizontal components of the vector.
X Component (horizontal): Cos(60)*40=20
Y Component (vertical): Sin(60)*40=34.64

Now that we know the vertical and horizontal components of vector A, we can add the corresponding components of B, although since we are ultimately subtracting B from A, we just make B negative.

This leaves us with:
X: 20 + (-50) = -30
Y: 34.64 + 0 = 34.64

Then we use the Pythagorean theorem to find the result vector of these new horizontal and vertical components
(A*A) + (B*B) = (C*C) = √(-30*-30) + (34.64*34.64) = C
C = 45.83

We know that the resultant vector has a magnitude of 45.83.
 
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