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Homework Help: EASY QUESTION supposedly

  1. Jan 30, 2009 #1
    Vector A has a magnitude of 2.00 m and is directed east. Vector B has a magnitude of 6.00 m and is directed 24.0° west of north.

    (1) What is the magnitude of A + B?
    (2) What is the direction of A + B?
    (3) What is the magnitude of B - A?
    (4) What is the direction of B - A?
  2. jcsd
  3. Jan 30, 2009 #2
    Break the vectors into their <x,y> components and then do the indicated operations.
  4. Jan 30, 2009 #3
    How do I break it? Do I use sine and cosine?
  5. Jan 30, 2009 #4
    Add two vectors. Find the x and y components of the two vectors. Add the x components to get the net x component. Do the same for the y components. Draw the net x component and the net y component. Magnitude is determined with the Pythagorean theorem, and angle is determined with an inverse tangent. If, instead, you add the vectors on a vector-capable calculator, all angles must be relative to the positive direction of the x axis.
  6. Jan 30, 2009 #5
    Yup. your first vector is already in component form because it is just pointed east, so its of the form <2,0>. (remember sign of the number matters in this).

    For your second one if you draw it out you'll notice you have a triangle. Use your sin and cos functions to find the X and Y components of the second vector.

    Then you'll have vector A broken into x and y =<x1,y1> and vector B broken into x and y <x2,y2> and then you can do the operations to them. Remember when you are doing the operations you can only add x to x and y to y.

    <1,1>+<2,5> = <3,6>

    [EDIT] and yes, about the angle with respect to the origin it brought up inverse tangent. That is because tan=(opp/adj)... which is pretty much the same thing in this case as tan(theta)=(y/x) so the angle w.r.t the positive x axis is theta=arctan(Y/X)
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