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Math Help - When are four points in R3 a parallelogram

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    When are four points in R3 a parallelogram

    Given P,Q,R,S in R3
    If one of PQ, PR, PS is equal to one of QR, QS, the points determine a parallelogram.
    Ex:
    P=(-1,0,2)
    Q=(3,4,-1)
    R=(3,2.-3)
    S=(-1.-2,0)

    PQ=(4,4,-3)
    PR=(4,2,-5)
    PS=(0,-2,-2)

    QR=(0,-2,-2), itís a parallelogram
    QS=(-4,-6,1)

    Since QR equals PS, it follows immediateley that QR and RS are adjacent sides of the parallelogram and the area is:

    A=|QRXRS|

    If you change S to (-1,-2,1), the conditions are not met and P,Q.R,S is not a parallelogram

    This proof did not show up on an Internet search, which is why I post it here.
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    Re: When are four points in R3 a parallelogram

    Quote Originally Posted by Hartlw View Post
    Given P,Q,R,S in R3
    P=(-1,0,2)
    Q=(3,4,-1)
    R=(3,2.-3)
    S=(-1.-2,0)

    If you change S to (-1,-2,1), the conditions are not met and P,Q.R,S is not a parallelogram

    This proof did not show up on an Internet search
    But that is exactly what I posted here.
    Quote Originally Posted by Plato View Post
    I find working with graph paper in \mathbb{R}^3 difficult. The four points may not even be co-planar . The four points determine six vectors. In order to have a parallelogram some pair of those six vectors must be parallel and have the same length. Recall that parallel vectors are multiples of each other.
    \\\overrightarrow {PQ}  = \left\langle {4,4, - 3} \right\rangle \\\overrightarrow {PR}  = \left\langle {4,2, - 5} \right\rangle \\\overrightarrow {PS}  = \left\langle {0,-2, - 2} \right\rangle \\\overrightarrow {QR}  = \left\langle {0, - 2, - 2} \right\rangle \\\overrightarrow {QS}  = \left\langle { - 4, - 6,1} \right\rangle \\\overrightarrow {RS}  = \left\langle { - 4, - 4,3} \right\rangle

    Note that \overrightarrow {PQ}  =  - \overrightarrow {RS} \;\& \;\left\| {\overrightarrow {PQ} } \right\| = \left\| {\overrightarrow {RS} } \right\| therefore, we have a parallelogram.

    Change S: (-1,-2,1) now
    \\\overrightarrow {PQ}  = \left\langle {4,4, - 3} \right\rangle \\\overrightarrow {PR}  = \left\langle {4,2, - 5} \right\rangle \\\overrightarrow {PS}  = \left\langle {0,-2, - 1} \right\rangle \\\overrightarrow {QR}  = \left\langle {0, - 2, - 2} \right\rangle \\\overrightarrow {QS}  = \left\langle { - 4, - 6,0} \right\rangle \\\overrightarrow {RS}  = \left\langle { - 4, - 4,-4} \right\rangle
    No two of those are parallel. So no parallelogram.

    With this nice geometric property, we can use this approach to identity of four points in \mathbb{R}^3 as being a rhombus or just a quadrilateral. That last point was my initial objection to your post in this thread.
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    Re: When are four points in R3 a parallelogram

    Quote Originally Posted by Hartlw View Post
    Given P,Q,R,S in R3
    If one of PQ, PR, PS is equal to one of QR, QS, the points determine a parallelogram.
    What, exactly, do you mean by "equal to"? Strictly speaking one line segment is "equal" to another if and only if they are the same line segment. And that's surely what NOT what you mean. I think you mean "if one of PQ, PR, PS is parallel to and has the same length as one of QR, QS".

    Ex:
    P=(-1,0,2)
    Q=(3,4,-1)
    R=(3,2.-3)
    S=(-1.-2,0)

    PQ=(4,4,-3)
    PR=(4,2,-5)
    PS=(0,-2,-2)
    Since you have use "(a, b, c)" to represent a point, I would be more comfortable with a different notation for the line segment. And, strictly speaking (again!) "(4, 4, 3)" (better notation <4, 4, 3> or 4i+ 4j+ 3k) is a vector parallel to the line, not the line or line segment itself.
    In terms of vectors, PQ= -1i+ 2k+ 4ti+ 4tj+ 3tk= (4t- 1)+ 4tj+ (3t+ 2)k for 0\le t\le 1, PR= -1i+ 2k+ 4ti+ 2tj- 5tk= (4t-1)I+ 2tj- (5t- 2)k, for 0\le t\le 1, and PS= -1i+ 2k- 2tj- 2tk= -ix- 2tj- (2t- 2)k, for 0\le t\le 1.

    QR=(0,-2,-2), itís a parallelogram
    Yes, QR and PS have the same "direction vector" and so they are both parallel and of the same length.[quote]
    Say that the vectors are equal, not the line segments, and you are correct. But that immediately equivalent to the fact that if two sides of a quadrilateral are parallel and of equal length then the quadrilateral is a parallelogram and that proof is in any secondary school geometry text.

    QS=(-4,-6,1)

    Since QR equals PS ,it follows immediateley that QR and RS are adjacent sides of the parallelogram and the area is:

    A=|QRXRS|

    If you change S to (-1,-2,1), the conditions are not met and P,Q.R,S is not a parallelogram

    This proof did not show up on an Internet search, which is why I post it here.
    I see no proof here, just a single example.
    Last edited by HallsofIvy; January 24th 2014 at 04:38 PM.
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    Re: When are four points in R3 a parallelogram

    Quote Originally Posted by Hartlw View Post
    Given P,Q,R,S in R3
    If one of PQ, PR, PS is equal to one of QR, QS, the points determine a parallelogram.
    Actually, this can be simplified to eliminate obvious mis-pairings:

    Given P,Q,R,S in R3
    If PR=+/-QS or PS=+/-QR, the points form a parallelogram.

    If PR=QS, PQ is adjacent to PR and area is PQXRS
    If PR=-QS=SQ, PS is adjacent to SQ and area is PSXSQ

    This is a simple and clean-cut result. You just calculate four vectors and check for equality. The proof is somewhat tedious.


    I agree your method is correct. It is easy to understand and the proof is trivial.


    PQ,QR, etc are obviously vectors.
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    Re: When are four points in R3 a parallelogram

    HallsofIvy
    Your method is simple and obvious, with the added benefit one knows exactly where it came from.

    However, as a programming algorithm, my method (apparently unique) would be much easier to implement: Calculate four vectors a, b, c, d and check a=b and c=d, rather than create a list of six vectors and compare them pair-wise for equality.

    But then, it's what the programmer feels most comfortable with.
    Last edited by Hartlw; January 25th 2014 at 12:11 PM. Reason: added "six"
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