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Homework Help: Vector Subtract Given Magnitude

  1. Dec 17, 2017 #1
    1. The problem statement, all variables and given/known data
    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.

    2. Relevant equations
    I'm given the magnitudes, and need to find magnitude of C

    3. The attempt at a solution
    I'm kinda confused cause I dont 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
     
  2. jcsd
  3. Dec 17, 2017 #2

    George Jones

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    Can you draw a diagram for the situation?
     
  4. Dec 17, 2017 #3
    upload_2017-12-17_23-38-33.png
     
  5. Dec 17, 2017 #4

    Delta²

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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?
     
  6. Dec 17, 2017 #5
    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

    [tex]c^2=a^2+b^2-2ab\cos{(0)}= (a-b)^2
    \\
    c = |a-b|

    [/tex]

    Which is what you tried to do but the angle is not zero in this case which is why it's wrong
     
  7. Dec 17, 2017 #6
    Thanks. I definitely did not learn this from my physics class.
     
  8. Dec 24, 2017 #7
    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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