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Math Help - veectors

  1. #1
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    veectors

    I have a question on vectors which probably doesn't belong here but I can't see an appropriate place to post it. If anyone has a better idea please tell me.

    Please see image attached:

    In the diagram, OABCDEFG is a cuboid in which the length of OA is 4 units, AB is 3 units and BF is 2 units. Unit vectors i, j, and k are parallel to OA, OC and OD respectively. The midpoints of AB and FG are M and N respectively.

    a) Express each of the vectors ON and MG in terms of i, j, and k.
    I would start this problem by finding ON→=OD→+DG→+1/2GN→
    I have place the arrow to the right of the signs as I can't put them above it.
    ON→=1/2i+j+k
    MG=1/2AB+BF+FG
    -i+1/2j+k
    Am I doing the right thing anyone?

    b) Find the acute angle between the direction of ON and MG, correct to the nearest 0.1.
    Will tackle this next.
    Attached Thumbnails Attached Thumbnails veectors-vectorbox020.jpg  
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  2. #2
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    Hello, stealthmaths!

    In the diagram, OABCD{E}FG is a cuboid
    . . in which: . |\overrightarrow{OA}| = 4,\; |\overrightarrow{AB}| = 3.\;|\overrightarrow{BF}| = 2 units.

    Unit vectors \vec i, \vec j, \vec k are parallel to \overrightarrow{OA}, \,\overrightarrow{OC}, \,\overrightarrow{OD}, respectively.

    The midpoints of AB and FG are M and N, respectively.

    \text{a) Express each of the vectors }\overrightarrow{ON} and \overrightarrow{MG}\text{ in terms of }\vec i,\, \vec j,\, \vec k.

    We see that: . \begin{array}{ccccc}<br />
\overrightarrow{OA} \:=\:\overrightarrow{CB} \:=\:\overrightarrow{DE} \:=\:\overrightarrow{GF} &=& 4i \\ <br />
\overrightarrow{OC} \:=\:\overrightarrow{AB} \:=\:\overrightarrow{DG} \:=\:\overrightarrow{EF} &=& 3j \\<br />
\overrightarrow{OD} \:=\:\overrightarrow{AE} \:=\:\overrightarrow{BF} \:=\:\overrightarrow{CG} &=& 2k <br />
\end{array}


    \overrightarrow{ON} \;=\;\overrightarrow{OC} + \overrightarrow{CG} + \tfrac{1}{2}\overrightarrow{GF} \;=\;3\vec j + 4\vec k + \tfrac{1}{2}(4\vec i)  \;=\;2\vec i + 3\vec j + 4\vec j


    \overrightarrow{MG} \;=\;\overrightarrow{MB} + \overrightarrow{BC} + \overrightarrow{CG} \;=\;\tfrac{1}{2}(3\vec j) - 4\vec i + 2\vec k \;=\;-4\vec i + \tfrac{3}{2}\vec j + 2\vec k




    \text{b) Find the acute angle between }\overrightarrow{ON} \text{ and }\overrightarrow{MG}\text{, to the nearest 0.1}^o.
    Angle \theta between \vec u and \vec v is given by: . \cos\theta \;=\;\frac{\vec u\cdot\vec v}{|\vec u||\vec v|}


    We have: . \overrightarrow{ON} \:=\:\langle 2,3,4\rangle,\;\;\overrightarrow{MG} \:=\:\langle -4,\tfrac{3}{2},2\rangle

    . . \overrightarrow{ON}\cdot \overrightarrow{MG} \:=\:\langle2,3,4\rangle \cdot\langle -4,\tfrac{3}{2},2\rangle \:=\:-8 + \tfrac{9}{2} + 8 \:=\:\tfrac{9}{2}

    . . |\overrightarrow{ON}| \:=\:\sqrt{4+9+16} \:=\:\sqrt{29}

    . . |\overrightarrow{MG}| \;=\;\sqrt{16 + \tfrac{9}{4} + 4} \;=\;\sqrt{\tfrac{89}{4}} \;=\;\frac{\sqrt{89}}{2}


    Hence: . \cos\theta \;=\;\frac{\frac{9}{2}}{\sqrt{29}\,\frac{\sqrt{89}  }{2}} \;=\;0.177152998

    Therefore: . \theta \;=\;79.79602628 \;\approx\;79.8^o

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  3. #3
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    Hi Soroban:
    Thanks for the reply:

    Soroban:
    On the first line of question a)
    ON=OC+CG+1/2CF=3j+4k+1/2(4i)
    Looking at your workings. Have you made a typo?
    Should it not be: 3j+2k+1/2(4i)?
    Last edited by stealthmaths; February 20th 2010 at 05:16 AM. Reason: posted statements in wrong post
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  4. #4
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    I have calculated with the correction to get the following:

    ON=(2, 3, 2)
    MG=(-4, 3/2, 2)

    ON.MG=(2, 3, 2)(-4, 3/2, 2)
    =-8+9/2+4=1/2

    |ON|=√4+9+4=√17
    |MG|=√16+9/4+4=√89/2

    hence: cosθ=1/2 / (√17.√89/2)=0.02570872484
    =88.5 (1d.p.)

    Solution found anyone?
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