When are four points in R3 a parallelogram

Aug 2010
961
101
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), it’s 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.
 

Plato

MHF Helper
Aug 2006
22,508
8,664
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.
I find working with graph paper in \(\displaystyle \mathbb{R}^3\) 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.
\(\displaystyle \\\overrightarrow {PQ} = \left\langle {4,4, - 3} \right\rangle \\\overrightarrow {PR} = \left\langle {4,2, - 5} \right\rangle \\\overrightarrow {PS} = \left\langle {0,-2, - 2} \right\rangle \\\overrightarrow {QR} = \left\langle {0, - 2, - 2} \right\rangle \\\overrightarrow {QS} = \left\langle { - 4, - 6,1} \right\rangle \\\overrightarrow {RS} = \left\langle { - 4, - 4,3} \right\rangle\)

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

Change \(\displaystyle S: (-1,-2,1)\) now
\(\displaystyle \\\overrightarrow {PQ} = \left\langle {4,4, - 3} \right\rangle \\\overrightarrow {PR} = \left\langle {4,2, - 5} \right\rangle \\\overrightarrow {PS} = \left\langle {0,-2, - 1} \right\rangle \\\overrightarrow {QR} = \left\langle {0, - 2, - 2} \right\rangle \\\overrightarrow {QS} = \left\langle { - 4, - 6,0} \right\rangle \\\overrightarrow {RS} = \left\langle { - 4, - 4,-4} \right\rangle\)
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 \(\displaystyle \mathbb{R}^3\) as being a rhombus or just a quadrilateral. That last point was my initial objection to your post in this thread.
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
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 \(\displaystyle 0\le t\le 1\), PR= -1i+ 2k+ 4ti+ 2tj- 5tk= (4t-1)I+ 2tj- (5t- 2)k, for \(\displaystyle 0\le t\le 1\), and \(\displaystyle PS= -1i+ 2k- 2tj- 2tk= -ix- 2tj- (2t- 2)k\), for \(\displaystyle 0\le t\le 1\).

QR=(0,-2,-2), it’s a parallelogram
Yes, QR and PS have the same "direction vector" and so they are both parallel and of the same length.
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.
 
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Aug 2010
961
101
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.
 
Aug 2010
961
101
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.
 
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