Vector Subtract Given Magnitude

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Homework Help Overview

The discussion revolves around finding the magnitude of vector C, defined as the difference between two vectors A and B, given their magnitudes and the angle between them. The problem is situated within the context of vector subtraction and involves concepts from vector mathematics, specifically the law of cosines and component analysis.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore different methods to find the magnitude of vector C, including the law of cosines and component breakdown of vectors. Questions arise regarding the correct application of these methods and the implications of the angle between the vectors.

Discussion Status

There is an ongoing exploration of the problem with various approaches being discussed. Some participants suggest using the law of cosines, while others consider component analysis. The original poster expresses confusion about the problem's complexity, and there is no explicit consensus on the correct method yet.

Contextual Notes

Participants note that the angle between vectors A and B is 60 degrees, which affects the calculations. There is also mention of a lack of understanding of the concepts from the original poster's physics class, indicating potential gaps in foundational knowledge.

Lori

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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