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Subtracting Vectors?

  1. Oct 27, 2009 #1
    I am completely lost. Here is a similar equation, I changed the lengths and angles.

    Vector A is 3.0 meters long, and is 45 degrees from the x axis.
    Vector B is 6.0 meters long, and is 130 degrees from the x axis.

    What are the x and y components of the vector C=B-A?
    What is the magnitude of C?
     
  2. jcsd
  3. Oct 27, 2009 #2
    First, draw a picture. Include a coordinate system and your vectors. Use trig (sine/cosine) to decompose your vectors into their x and y components.

    C=B-A is a vector equation, so Cx=Bx-Ax and Cy=By-Ay
     
  4. Oct 27, 2009 #3
    I'm assuming you know how to add vectors, but if not, replay back.

    Subtracting vectors is the same thing as adding vectors, but in the opposite direction.

    So, in your example, subtracting A would be the same thing as adding 3.0 meters 225 degrees from the x-axis.
     
  5. Oct 27, 2009 #4
    I don't know how to add vectors at all. I know I have to use cos and sin but my knowledge ends there.
     
  6. Oct 27, 2009 #5
    Just focus on vector A right now, sketch a coordinate system and draw in the vector with the angle you described. Sin(the angle)=what? Cos (the angle)=what? You can use the sine and cosine function to get the x and y components of A.

    Vectors are nothing but triangles, don't worry about them.
     
  7. Oct 27, 2009 #6
    Thank you so much for your help. What about the lengths? So far for that problem, I have:

    cos (45)= .707
    sin (45)= .707
    cos (130)= -0.643
    sin (130) = 0.766

    I just don't have any clue what to do with those numbers.
     
  8. Oct 27, 2009 #7
    So what you have calculated is the ratio of the length (magnitude) of your vectors to their x and y components. For vector A you have sin(45 deg)=Ay/A so that Ay=A*sin(45 deg)=3*sin(45deg)=0.707*sin(45 deg). For the x component Ax=A*cos(45 deg).

    Now you have the x and y components of your A vector, what are the x and y components of your B vector?
     
  9. Oct 27, 2009 #8
    I got:

    Ax= 2.121
    Ay= 2.121
    Bx= -3.858
    By= 4.596

    Thank you so much for your help! How do I subtract them? :)
     
  10. Oct 27, 2009 #9
    I think I may have figured it out.

    Cx= -5.979
    Cy= 2.475

    and the Magnitude= 6.471

    Thank you sooo much!! I am so appreciative!! :D You are my savior!!
     
  11. Oct 27, 2009 #10
    You're doing great. Remember your original equation for C? C=B-A which was a vector equation which really means it is two equations in one. Cx=Bx-Ax so Cx=-3.858-2.121 and similarly for Cy. Now remember the problem originally asked for the magnitude of the vector C but since Cx and Cy are perpendicular to each other you can use the Pythagorean theorem to find the total length of C. And that should do it.

    I know this vector stuff must seem sort of weird and abstract now but I think once you apply it to some real problems it will make a little more sense and eventually be more intuitive.
     
  12. Oct 27, 2009 #11
    Thank you so much! I did sqrt(cx^2+cy^2)=C. Is that right? For magnitude, that is.
     
  13. Oct 27, 2009 #12
    Perfect :)
     
  14. Oct 27, 2009 #13
    Thank you so much!! :d
     
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