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Math Help - Vector's (proving 1 line is parallel to another)

  1. #1
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    Vector's (proving 1 line is parallel to another)

    I have this question that I am really stuck with, I have done the first part and believe I am close to solving the second part but really cannot see what to do next.

    Q) ABCD is a trapezium with BC parallel to AD.
    AB=3b BC=3a AD=9a
    M is the midpoint of BC and N is the midpoint of AD
    a) Find the vector MN in terms of a and b - which i found to be 3a-3b

    b) Prove that XY is parallel to AD

    if anybody could help me with this part it would be much appreciated

    Thanks
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  2. #2
    Senior Member BAdhi's Avatar
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    What is XY here?
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  3. #3
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    sorry missed out 1 line of the question here it is again:

    I have this question that I am really stuck with, I have done the first part and believe I am close to solving the second part but really cannot see what to do next.

    Q) ABCD is a trapezium with BC parallel to AD.
    AB=3b BC=3a AD=9a
    M is the midpoint of BC and N is the midpoint of AD
    a) Find the vector MN in terms of a and b - which i found to be 3a-3b

    X is the midpoint of MN and Y is the midpoint of CD.
    b) Prove that XY is parallel to AD

    if anybody could help me with this part it would be much appreciated

    Thanks
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  4. #4
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    \\* \vec{MN} = 3a - 3b \\* \vec{CD} = (-3a - 3b + 9a) = 6a - 3b

    X and Y are the mid-points of MN and CD, respectively.
    \\* X = \frac{1}{2}\vec{MN} \\* Y = \frac{1}{2}\vec{CD}

    Subtract the two midpoint vectors to get the resulting vector.
    \\* \vec{XY} = Y - X \\*  = \frac{1}{2}(6a - 3b) - \frac{1}{2}(3a - 3b)  \\*  = \frac{1}{2}(3a)

    As you already know:
     \\* \vec{AD} = 9a

    Both vectors consist of only 'a' vectors, so they are parallel.
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  5. #5
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    Hello, wahhdoe!

    I'll solve it in my usual baby-talk way . . .


    ABCD\text{ is a trapezium with }BC \parallel AD.
    \overrightarrow{AB} = 3\vec b,\;\; \overrightarrow{BC} = 3\vec a,\;\; \overrightarrow{AD} = 9\vec a
    M\text{ is the midpoint of }BC\text{ and }N\text{ is the midpoint of }AD.

    \text{a) Find the vector }\overrightarrow{MN}\text{ in terms of }\vec a\text{ and }\vec b
    . . \text{-- which I found to be }3\vec a-3\vec b . Yes!

    X\text{ is the midpoint of }MN\text{ and }Y\text{ is the midpoint of }CD.

    \text{b) Prove that }XY \parallel AD
    Code:
                : - -  3a - - :
                      M
              B o * * o * * o C
               *       *       *
              *         *         *   Y
          3b *         X o * * * * * o
            *             *             *
           *               *               *
        A o  *  *  *  *  *  o  *  *  *  *  *  o D
                            N
          : - - - - - - - -9a - - - - - - - - :

    (a)\;\overrightarrow{MN} \:=\:\overrightarrow{MB} + \overrightarrow{BA} + \overrightarrow{AN}

    . . . . . . . =\;\text{-}\tfrac{1}{2}\overrightarrow{BC} - \overrightarrow{AB} + \tfrac{1}{2}\overrightarrow{AD}

    . . . . . . . =\;\text{-}\tfrac{1}{2}(3\vec a) - (3\vec b) + \tfrac{1}{2}(9\vec a)

    . . . . . . . =\;\text{-}\tfrac{3}{2}\vec a - 3\vec b + \tfrac{9}{2}\vec a

    . . . . . . . =\;3\vec a - 3\vec b



    (b)\;\overrightarrow{CD} \;=\;\overrightarrow{CB} + \overrightarrow{BA} + \overrightarrow{AD}

    . . . . . . =\;-3\vec a - 3\vec b + 9\vec a

    . . . . . . =\;6\vec a - 3\vec b


    \overrightarrow{XY} \;=\;\overrightarrow{XM} + \overrightarrow{MC} + \overrightarrow{CY}

    . . . . =\;\text{-}\tfrac{1}{2}\overrightarrow{MN} + \tfrac{1}{2}\overrightarrow{BC} + \tfrac{1}{2}\overrightarrow{CD}

    . . . . =\;\text{-}\tfrac{1}{2}(3\vec a - 3\vec b) + \tfrac{1}{2}(3\vec a) + \tfrac{1}{2}(6\vec a - 3\vec b)

    . . . . =\;\text{-}\tfrac{3}{2}\vec a + \tfrac{3}{2}\vec b + \tfrac{3}{2}\vec a + 3\vec a - \tfrac{3}{2}\vec b

    . . . . =\; 3\vec a


    \text{We have: }\:\overrightarrow{XY} \:=\:3\vec a \:=\:\tfrac{1}{3}\overrightarrow{AD}

    \text{Therefore: }\:\overrightarrow{XY} \parallel \overrightarrow{AD}

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