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Angle between Vector question

  1. Jan 19, 2007 #1
    1. The problem statement, all variables and given/known data
    Two vectors A and B have precisely equal magnitudes. In order for the magnitude of A + B to be 100 times larger than the magnitude of A - B, what must be the angle between them?

    2. Relevant equations

    3. The attempt at a solution

    I am completely lost. Vectors are a completely new subject to me. I am not familiar with vector algebra notation yet as we haven't yet covered it, but I assume we are to solve this visually/conceptually given what we do know. Any help where to start?
  2. jcsd
  3. Jan 19, 2007 #2


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    HINT: The scalar product between A+B and A-B is 0, since the 2 vectors are perpendicular.

  4. Jan 19, 2007 #3
    Hrm maybe if I had a step by step explanation I would understand what was going on. I didn't even know what "vector" was yesterday. I appreciate the hint daniel, but I still have no idea.
  5. Jan 19, 2007 #4
    Sorry... A and B are not perpendicular


    Hints: |A+B| = 100 |A-B|

    notice |A+B|^2 = (A+B) dot (A+B) and
    A dot B = |A| |B| cos \theta
    Last edited: Jan 19, 2007
  6. Jan 19, 2007 #5


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    This seems backwards: actually A+B and A-B are perpendicular (i.e. they have scalar product zero) because A and B are equal magnitude.

    Moonworm, if you are just starting vectors, draw a diagram with two vectors from a point, with equal magnitude, and an angle theta between them. Then draw the vectors a+b (using the parallelogram rule) and a-b.

    You can find the lengths of a+b and a-b using geometry and trig.
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