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Math Help - vector parallelogram

  1. #1
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    vector parallelogram

    ABCD is a parallelogram. A, B and C are represented by the position vectors
    i + 2j - k, 2i + j -2k and 4i - k respectively. Find:
    a) A->D
    b) The cosine of <BAD
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  2. #2
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    Hi, have a look at the attached diagrams:
    Attached Thumbnails Attached Thumbnails vector parallelogram-vect_parallelog1.gif   vector parallelogram-vect_parallelog2.gif  
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  3. #3
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    Hello, scorpion007!

    ABCD is a parallelogram.
    A, B and C are represented by the position vectors:
    . . i + 2j - k,\;2i + j -2k\;4i - k, respectively.

    (a)\text{ Find: }\overrightarrow{AD}

    We have: . A = \langle1,2,\text{-}1\rangle,\;B = \langle2,1,\text{-}2\rangle,\;C = \langle4,0,\text{-}1\rangle,\;D = \langle x,y,z\rangle

    Then: . \overrightarrow{BA} = \langle\text{-}1,\,1,<br />
\,1\rangle and \overrightarrow{CD} = \langle x-4,\,y-0,\,z+1\rangle

    Since BA \parallel CD:\;\begin{array}{ccc}x - 4 \,=\,\text{-}1 \\ y\,=\,1 \\ z+1\,=\,1\end{array} . \Rightarrow . \begin{array}{ccc}x=3 \\ y=1 \\ z = 0\end{array}

    Therefore: . D = \langle3,1,0\rangle\quad\Rightarrow\quad AD \:=\:\langle3-1,\,1-2,\,0-(\text{-}1)\rangle\:=\:\langle2,\text{-}1,1\rangle



    (b)\text{ Find }\cos(\angle BAD)

    The angle \theta between two vectors \vec{u} and \vec{v} is given by: . \cos\theta \:=\:\frac{\vec{u}\cdot\vec{v}}{|\vec{u}||\vec{v}|  }

    We have: . \begin{array}{cc}\overrightarrow{AB} = \langle1,\text{-}1,\text{-}1\rangle \\ \overrightarrow{AD} = \langle2,\text{-}1,1\rangle\end{array} . \Rightarrow . \begin{array}{cc}|AB| = \sqrt{3} \\ |AD| = \sqrt{6}\end{array}

    . . and: . \overrightarrow{AB}\cdot\overrightarrow{AD} \:=\:(1)(2)+(\text{-}1)(\text{-}1)+(\text{-}1)(1) \:=\:2

    Therefore: . \cos(\angle BAD) \;= \;\frac{2}{\sqrt{3}\sqrt{6}}\;=\;\frac{\sqrt{2}}{3  }

    . . \left(\theta \:\approx\:61.87^o\right)

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  4. #4
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    thanks for the help guys
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  5. #5
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    earboth, according to your diagram, isn't a + BC = d or OD? not AD as you have it written?
    since AD || BC, and since its a parallelogram, shouldnt AD = BC? so why is AD = a + BC? Im thinking a + BC yields the vector from the origin to D, OD.

    EDIT: i see [.math] notation is back.
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  6. #6
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    Quote Originally Posted by scorpion007 View Post
    earboth, according to your diagram, isn't a + BC = d or OD? not AD as you have it written?
    since AD || BC, and since its a parallelogram, shouldnt AD = BC? so why is AD = a + BC? Im thinking a + BC yields the vector from the origin to D, OD....
    Hi,

    you are right. Sorry for that typo.

    EB
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