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), 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.

Re: When are four points in R3 a parallelogram

Quote:

Originally Posted by

**Hartlw** 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** 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.

Re: When are four points in R3 a parallelogram

Quote:

Originally Posted by

**Hartlw** 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".

Quote:

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 $\displaystyle 0\le t\le 1$.

Quote:

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.[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.

Quote:

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.

Re: When are four points in R3 a parallelogram

Quote:

Originally Posted by

**Hartlw** 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.

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.