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The vector resolute.Why doesn't it cancel out?

  1. Sep 15, 2008 #1
    Hello everyone :smile: ,

    I'm reviewing vectors and I thought the best way to go about it is understand the exact definition without blindingly using formulas. Anyway here is my question

    The vector resoulute of a on b

    * - unit vector

    (a . b*) b*

    My question is why doesn't this cancel out to a like this.

    a. b x b
    ----- ---
    |b| |b|

    |b|2 = b.b

    So wouldn't this be

    a.b x b
    which is equal to a

    I know this doesn't happen but I think there is something wrong in my understanding. I can't understand what happens when two vectors are multiplied. Why does it give a scalar. How do you explain it. Can you multiply 3 vectors. Can you divide vectors?

    It seems a lot of questions but I think by answering my first question you would probably answer these questions. Thanks a lot in advance :smile:
  2. jcsd
  3. Sep 15, 2008 #2


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    Hi sameeralord! :smile:

    You're asking why doesn't (a.b)b/b2 = a.

    Because (a.b)b isn't a(b.b) …

    the first one is a scalar multiple of b, and the second is a scalar multiple of a.
    You can either dot-product two vectors, or cross-product them:

    a.b and a x b.

    the first is a scalar, the second is a vector.

    Why? Because that's how we define them, and we can define anything we like.

    You can't divide vectors unless one is a scalar times the other (in other words, they're parallel).

    You can't multiply more than two vectors, except that you can keep cross-producting vectors in pairs, as many times as you like:

    ((axb)xc)xd etc,

    and you can also do the scalar "triple product" (axb).c :smile:
  4. Sep 15, 2008 #3
    Thanks heaps tiny tam :smile: Your answer was very clear. So basically you can't apply all the algebra techniques to vectors. I mean if there is (b.b) you take it as a scalar then do your calculations. So |b2| = (b.b) is also a scalar then. Thanks a lot again :smile:
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