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Vector question

  1. Aug 9, 2009 #1
    Hi All,

    I'm currently studying vector projections and the vector dot product. I ran into a problem on the homework I wasn't quite sure how to tackle..... Any suggestions?


    The problem is stated as follows:

    Determine ||v+w|| if v and w are unit vectors seperated by an angle of [tex]\frac{\pi}{6}[/tex]

    Thanks!

    Josh
     
  2. jcsd
  3. Aug 9, 2009 #2

    Hootenanny

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    Welcome to Physics Forums.

    HINT: Can you express the magnitude of the sum as a series dot products?
     
  4. Aug 9, 2009 #3
    Thanks.

    Hmm I don't follow you.
     
  5. Aug 9, 2009 #4

    Hootenanny

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    What is the definition of the magnitude of a vector, in terms of scalar products?
     
  6. Aug 9, 2009 #5

    Well I know the scalar product(dot product) is defined to be:

    u*v=||u||*||v||cos[tex]\theta[/tex]

    I've been lookin at it for the past hour or so and can't put it together how this definition applies.
     
  7. Aug 9, 2009 #6

    Hootenanny

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    Well you also know that for any vector, a,

    ||a|| = (a.a)1/2

    Does that help?
     
  8. Aug 9, 2009 #7
    Not really, I just totally don't see the direction your going in.... I don't have any information about the vector's components.
     
  9. Aug 9, 2009 #8

    Hootenanny

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    Try explicity computing the magnitude of the sum, i.e.

    ||u + w|| = ...
     
  10. Aug 9, 2009 #9
    Sorry Hootenanny, I'm lost....


    Josh
     
  11. Aug 9, 2009 #10

    VietDao29

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    Ok, so here's the main formula that we'll use to solve the problem.

    u, and v can be any vector. So, if u = v = a, that means the angle [tex]\theta[/tex] between those 2 vectors is 0, right? Both u, and v are the same vector, so there's no angle separating them.

    Using the formula above, we have:

    [tex]\mathbf{a} * \mathbf{a} = ||\mathbf{a}|| ||\mathbf{a}|| \cos (0)[/tex]

    [tex]\Rightarrow \mathbf{a} * \mathbf{a} = ||\mathbf{a}|| ^ 2[/tex] (since cos(0) = 1)

    [tex]\Rightarrow ||\mathbf{a}|| = \sqrt{\mathbf{a} * \mathbf{a}}[/tex], which boils down to the formula Hoot showed you:

    Now, since u, and v are both vectors, u + v is also a vector. How can we determine ||u + v||. How can we apply the above formula?

    Let's assume that you can get over this step. After applying the formula, you will arrive at some expression. The next step is to expand it. Then use the first formula

    [tex]\mathbf{u} * \mathbf{v} = ||\mathbf{u}|| ||\mathbf{v}|| \cos (\theta)[/tex]

    to simplify that expression, and arrive at the desired result. Also note that w, and v in your problem are unit vectors, i.e: ||u|| = ||v|| = 1

    This one is very basic, sorry I cannot be more detailed, or otherwise, I'll be solving it for you. Hopefully you can go from here, right? :)
     
  12. Aug 9, 2009 #11
    Josh, I wouldn't use the dot product at all. I'd break both vectors into components (x and y parts), add the components, then find the magnitude of the resultant using the Pythagorean theorem.

    Pick your x-axis along the same direction as W. That way, W only has an x-component (it makes life a little easier). Since V makes a 30-degree angle from it, can you calculate what the x and y-components of V would be?
     
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