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Math Help - Two vector problems I need some help with

  1. #1
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    Two vector problems I need some help with

    1. Determine whether the vectors [2,-1] and [-2,-5] are orthogonal.

    I remember learning this stuff last year and I think it has something to do with the dot product of vectors maybe? Help would be greatly appreciated.

    2. Find the angle between vector u = [-2, 5] and vector v= [-1, 3].

    Once again, I completely forget how to do this. Cross product?
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  2. #2
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    Quote Originally Posted by Jph93 View Post
    1. Determine whether the vectors [2,-1] and [-2,-5] are orthogonal.
    2. Find the angle between vector u = [-2, 5] and vector v= [-1, 3].
    If the dot product is zero, [2,-1]\cdot [-2,-5]=? they are orthogonal.

    The angle between u~\&~v is \arccos \left( {\frac{{u \cdot v}}{{\left\| u \right\|\left\| v \right\|}}} \right)
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  3. #3
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    So -4+5= 1
    not orthogonal

    and

    17/(5.39)(3.16)
    = 17/17
    =1
    arccos1=0

    so 0 degrees?
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  4. #4
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    Quote Originally Posted by Jph93 View Post
    17/(5.39)(3.16)
    = 17/17
    =1
    arccos1=0
    You have missed the lengths.
    \left\| u \right\| = \sqrt {29} \;\& \,\left\| v \right\| = \sqrt {10}
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    They equal 17 when multiplied right?
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  6. #6
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    Quote Originally Posted by Jph93 View Post
    They equal 17 when multiplied right?
    Oh NO!
    \sqrt {10}  \cdot \;\sqrt {29}  \ne 17\;,\,17^2  = 289 \ne 290
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  7. #7
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    Quote Originally Posted by Plato View Post
    If the dot product is zero, [2,-1]\cdot [-2,-5]=? they are orthogonal.

    The angle between u~\&~v is \arccos \left( {\frac{{u \cdot v}}{{\left\| u \right\|\left\| v \right\|}}} \right)
    I am still re-learning my Algebra and have not started on pre-calc yet, but from what I remember about vectors, this notation [2,-1] is the distance of the vector from the origin in the x and y directions, correct?

    If this is the case, could you not also solve this problem by calculating then comparing the slope of both vectors as lines?

    ie,  m = \frac{y_2 - y_1}{x_2 - x_1},  \,

      m_1 =  \frac{-1 - 0}{2 - 0} = \frac{-1}{2}, \,

      m_2 =  \frac{-5 - 0}{-2 - 0} = \frac{5}{2}

    since    m_1 \neq  \frac{-1}{m_2} , vectors are not orthogonal.

    Is this approach correct?
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  8. #8
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    Well, that is correct as far as it goes.
    But notice that the OP is about vectors, whereas you have used traditional analytical geometry.
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  9. #9
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    Quote Originally Posted by Plato View Post
    If the dot product is zero, [2,-1]\cdot [-2,-5]=? they are orthogonal.

    The angle between u~\&~v is \arccos \left( {\frac{{u \cdot v}}{{\left\| u \right\|\left\| v \right\|}}} \right)
    First one not orthogonal because the dot product is not 0?

    So Plato, is the answer for the angle one 0.05875 radians or 3.37 degrees?

    Ohh I know this ain't my problem but I stumbled across it and found it good practise on recapping my knowledge on these things...
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  10. #10
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    Quote Originally Posted by dwat View Post
    , is the answer for the angle one 0.05875 radians or 3.37 degrees?
    That is approx correct.
    If the angle between two vectors is 0\text{ or }\pi then the vectors must be multiples of one another.
    That is clearly not the case here.
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