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Math Help - Vectors 3

  1. #1
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    Vectors 3

    If a and b are unit vectors, and |a-b|=5, evaluate (2a+5b) . (3a-4b).

    Ok, so can someone please check my work for me? I am just not sure on one part, finding the angle of cos(theta). Btw this is the dot product.

    a and b both have u-hats on them

    6(a)^2 + 7(a)(b)-20(b)^2
    =6(1)^2+7(0.5)-20(1)^2
    =-10.5

    dot product work:
    a . b = |a||b|cos(theta)
    =(1)(1)cos60
    =0.5

    I was wondering if I had my angle theta correct
    To get it, I used the cos law,
    so theta is the angle across from 1 in my diagram.. am I supposed to look for the angle between a and b or between a and a-b.. Because I looked for the one between a and a-b
    cos(theta)=(1+5-1)/(2(1)(5))
    theta=0.5
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  2. #2
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    You are far off on this one.
    \begin{gathered}<br />
  25 = \left( {a - b} \right) \cdot \left( {a - b} \right) = a \cdot a - 2a \cdot b + b \cdot b \hfill \\<br />
  23 =  - 2a \cdot b \hfill \\ \end{gathered}

    \left( {2a + 5b} \right) \cdot \left( {3a - 4b} \right) = 6a \cdot a + 7a \cdot b - 20b \cdot b

    You finish it off.
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  3. #3
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    just to confirm, because i dont see the u-hat on your a and b.. your work involves a and b as unit vectors, correct?
    and how did you get from |a|^2 -2 a. b +|b|^2
    to the line 23=-2a.b
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  4. #4
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    Quote Originally Posted by skeske1234 View Post
    just to confirm, because i dont see the u-hat on your a and b.. your work involves a and b as unit vectors, correct?
    and how did you get from |a|^2 -2 a. b +|b|^2
    to the line 23=-2a.b
    25 = \left\| {a - b} \right\|^2  = \left( {a - b} \right) \cdot \left( {a - b} \right) = a \cdot a - 2a \cdot b + b \cdot b= 1 - 2a \cdot b + 1

    BTW: Hats are for birthday parties.
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  5. #5
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    Ok, thanks! just one more question though, when you go from the 2nd step to the 3rd step, I'm wondering if there is a dot product rule or property that you are supposed to use to know that you have to seperate the squared into (a-b) (a-b), or, did you just know that you had to do that, in order too find a.b
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  6. #6
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    Quote Originally Posted by skeske1234 View Post
    If a and b are unit vectors, and |a-b|=5, evaluate (2a+5b) . (3a-4b).
    Now correct me if I'm wrong, but isn't it the case that |a-b|\leq |a|+|b|=2?

    So then how can it be that |a-b|=5?

    Shurely shome mishtake?
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  7. #7
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    Every a and b in the question is a unit vector so..
    every a and b in the question has a u-hat over-top of it..

    is Plato's method wrong?
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  8. #8
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    Quote Originally Posted by skeske1234 View Post
    I'm wondering if there is a dot product rule or property that you are supposed to use to know that you have to seperate the squared into (a-b) (a-b), or, did you just know that you had to do that, in order too find a.b
    Is this what you mean?
    \begin{gathered}<br />
  \left( {a - b} \right) \cdot \left( {a - b} \right) = a \cdot a - 2a \cdot b + b \cdot b \hfill \\<br />
  \left( {a + b} \right) \cdot \left( {a - b} \right) = a \cdot a + b \cdot b \hfill \\<br />
  \left( {a + b} \right) \cdot \left( {a + b} \right) = a \cdot a + 2a \cdot b + b \cdot b \hfill \\ <br />
\end{gathered}
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  9. #9
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    Quote Originally Posted by Plato View Post
    Is this what you mean?
    \begin{gathered}<br />
\left( {a - b} \right) \cdot \left( {a - b} \right) = a \cdot a - 2a \cdot b + b \cdot b \hfill \\<br />
\left( {a + b} \right) \cdot \left( {a - b} \right) = a \cdot a + b \cdot b \hfill \\<br />
\left( {a + b} \right) \cdot \left( {a + b} \right) = a \cdot a + 2a \cdot b + b \cdot b \hfill \\ <br />
\end{gathered}
    that and the step before that
    (a-b)^2 = (a-b) . (a-b)
    is there a rule or property of the dot product that says you need to or can split the squared into two, or do you just do that from normal mathematical approach?

    also with
    \begin{gathered}<br />
\left( {a - b} \right) \cdot \left( {a - b} \right) = a \cdot a - 2a \cdot b + b \cdot b \hfill \\<br />
\left( {a + b} \right) \cdot \left( {a - b} \right) = a \cdot a + b \cdot b \hfill \\<br />
\left( {a + b} \right) \cdot \left( {a + b} \right) = a \cdot a + 2a \cdot b + b \cdot b \hfill \\ <br />
\end{gathered}
    are these three too, rules or property of the dot product as well?
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  10. #10
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    Quote Originally Posted by halbard View Post
    Now correct me if I'm wrong, but isn't it the case that |a-b|\leq |a|+|b|=2?

    So then how can it be that |a-b|=5?

    Shurely shome mishtake?
    No, that is correct.
    I never even noticed that.
    So the question itself is wrong.
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  11. #11
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    Quote Originally Posted by Plato View Post
    25 = \left\| {a - b} \right\|^2  = \left( {a - b} \right) \cdot \left( {a - b} \right) = a \cdot a - 2a \cdot b + b \cdot b= 1 - 2a \cdot b + 1
    Hello.
    Hello
    Hello

    Is anybody listening?

    The question is inconsistent.

    Given the above, we have \cos\theta=\frac{a\cdot b}{|a||b|}=-\frac{23}2. I'm dying to know which angle this is.

    Please check the question. There is a mistake somewhere. If a and b are unit vectors then |a-b|=5 is impossible.
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  12. #12
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    then...... did we do a.b correct?
    Last edited by skeske1234; August 27th 2009 at 04:12 PM.
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  13. #13
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    Quote Originally Posted by skeske1234 View Post
    then...... did we do a.b correct?
    The problem is wrong.
    Therefore, there is no way to do a\cdot b correctly.
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