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    Q:

    Points A, B and C have coordinates (5, -1, 0), (2, 4, 10) and (6, -1, 4) respectively.

    (a) Find the vectors \vec{CA} and \vec{CB}

    (b) Find the area of the triangle ABC

    (c) Point D is such that the point A, B, C and D form the vertices of a parallelogram. Find the coordinates of three possible positions of D.

    --------

    (a) \vec{CA} = \begin{pmatrix} -1 \\ 0 \\ -4 \end{pmatrix}

    \vec{CB} = \begin{pmatrix} -4 \\ 5 \\ 6 \end{pmatrix}

    (b) Area of triangle = 15.07 units^2

    (c) Need help in this part.

    Zacken any help with part (c) please?
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    What's the definition of parallelogram?
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    (Original post by Zacken)
    What's the definition of parallelogram?
    A four sided shape, with opposite sides parallel I guess?
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    (Original post by SaadKaleem)
    A four sided shape, with opposite sides parallel I guess?
    Yep, so set up your D such that CD is parallel to AB or DA is parallel to another side or etc...
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    (Original post by Zacken)
    Yep, so set up your D such that CD is parallel to AB or DA is parallel to another side or etc...
    Yes, I managed to sketch three possible shapes, however failed to deduce regardless, the coordinates of D.

    Probably missing out on something silly here.

    Vectors :argh:

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    (Original post by SaadKaleem)
    Yes, I managed to sketch three possible shapes, however failed to deduce regardless, the coordinates of D.

    Probably missing out on something silly here.

    Vectors :argh:
    Set D = (x, y, z)^{T} and then look at AD, CD, BD, or along those lines and equate AD with the opposite side of AD - since it's a parallelogram, then compare components, etc...
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    (Original post by Zacken)
    Set D = (x, y, z)^{T} and then look at AD, CD, BD, or along those lines and equate AD with the opposite side of AD - since it's a parallelogram, then compare components, etc...
    Alright, got my answers.. This was my first shape-like vector question, so i struggled.

    Answers I got:

    \vec{OD 1} = \begin{pmatrix} 9 \\ -6 \\ -6 \end{pmatrix}

    \vec{OD 2} = \begin{pmatrix} 1 \\ 4 \\ 6 \end{pmatrix}

    \vec{OD 3} = \begin{pmatrix} 3 \\ 4 \\ 14 \end{pmatrix}

    So in coordinates (9, -6, -6), (1, 4, 6) and (3, 4, 14)

    Thank you Zacken
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    (Original post by SaadKaleem)
    Thank you Zacken
    First class work!
 
 
 
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