# Thread: vectors - prove shape is a parallelogram

1. ## Re: vectors - prove shape is a parallelogram

Originally Posted by Plato
The statement that two of $\overrightarrow {PQ} ,\;\overrightarrow {PS} \;\overrightarrow {PR}$ "have to be adjacent sides of a quadrilateral" with vertices $P,~Q,~R,~S$ is false.
That is not what I said. This is what I said:

“1) Find PQ,PR,PS: two of these have to be adjacent sides of a quadrilateral.”

I obviously meant a space (non-planar) quadrilateral since we are talking about points in R3. quadri-four, lateral-side.

However, if that bothers you, Any two lines emanating from a point are sides of some (plane) quadrilateral(s): by elementary geometry.

Either one works.

I find it hard to believe that this has to be explained to you.

2. ## Re: vectors - prove shape is a parallelogram

Originally Posted by Hartlw
I obviously meant a space (non-planar) quadrilateral since we are talking about points in R3. quadri-four, lateral-side.
You have just made my point for me. There is no such concept as a " a space (non-planar) quadrilateral".

This question has appeared in most of the vector approach to geometry text I have reviewed.
It is usually more carefully worded: Given four points, no three of which are colinear, six vectors can be determined. If any pair of those vectors are parallel and have same length then the four points are the vertices of a parallelogram.

Originally Posted by Hartlw
I find it hard to believe that this has to be explained to you.
That is my sentiment about your contribution exactly.

3. ## Re: vectors - prove shape is a parallelogram

Originally Posted by Hartlw
Without any pre-conceptions of where P,Q,R,S are:

1) Find PQ,PR,PS: two of these have to be adjacent sides of a quadrilateral.
2) Find QR,QS: One of these has to be a side other than the two adjacent sides of 1)

If one of 1) is equal to one of 2) it’s a parallelogram.

Sounds good, but really not intelligible (to me) without playing with some sketches of four arbitrary points.
Perhaps some examples might help:

From the OP:
P=(-1,0,2)
Q=(3,4,-1)
R=(3,2.-3)
S=(-1.-2,0)
1)
PQ=(4,4,-3)
PR=(4,2,-5)
PS=(0,-2,-2)
2)
QR=(0,-2,-2), it’s a parallelogram

Furthermore, it follows that:
PQ and QR are adjacent sides of the parallelogram so:
A=|PQXQR|

How about a non-planar example:
P= (0,0,0)
Q=(1,0,0)
R=(0,1,0)
S=(0,0,1)
1)
PQ=(1,0,0)
PR=(0,1,0)
PS=(0,0,1)
2)
QR=(-1,1,0)
QS=(-1,0,1)
it’s not a parallelogram

No guesswork, no sketch, no pre-conceptions (about where the points are) required.

4. ## Re: vectors - prove shape is a parallelogram

Given any P,Q,R,S in R3,
If one of QR, QS is equal to one of PQ, PR, PS, the points determine a parallelogram.

See previous post for examples.

5. ## Re: vectors - prove shape is a parallelogram

Originally Posted by Plato
The statement that two of $\overrightarrow {PQ} ,\;\overrightarrow {PS} \;\overrightarrow {PR}$ "have to be adjacent sides of a quadrilateral" with vertices $P,~Q,~R,~S$ is false.
Of those three vectors, any two of them are co-planar (because three points determine a plane) and so are adjacent sides of some quadrilateral (in fact, some parallelogram). That appears to be all that was said.

6. ## Re: vectors - prove shape is a parallelogram

Originally Posted by HallsofIvy
Of those three vectors, any two of them are co-planar (because three points determine a plane) and so are adjacent sides of some quadrilateral (in fact, some parallelogram). That appears to be all that was said.
That is what I originally said:

"Without any pre-conceptions of where P,Q,R,S are:

1) Find PQ,PR,PS: two of these have to be adjacent sides of a quadrilateral."

Why is it repeated over and over and over again? That's why, out of shear frustration, I reworded the solution as:

Given any P,Q,R,S in R3,
If one of QR, QS is equal to one of PQ, PR, PS, the points determine a parallelogram.

7. ## Re: vectors - prove shape is a parallelogram

Originally Posted by HallsofIvy
Of those three vectors, any two of them are co-planar (because three points determine a plane) and so are adjacent sides of some quadrilateral (in fact, some parallelogram). That appears to be all that was said.
Actually here from reply #7 is the contested quote:
Originally Posted by Hartlw
Without any pre-conceptions of where P,Q,R,S are:
1) Find PQ,PR,PS: two of these have to be adjacent sides of a quadrilateral.
2) Find QR,QS: One of these has to be a side other than the two adjacent sides of 1)
.
Surely that is calming that $P,~Q,~R,~\&~S$ are co-planar?

8. ## Re: vectors - prove shape is a parallelogram

Theorem:
Four points P,Q,R,S in R3, regardless of labelling, form a parallelogram if
PS=±RQ or PQ=±RS

Proof:
If you take PQ,PR,PS in pairs and place the remaining point to form a parallelogram, you get requirements on pairs of points. Eliminating the redundant ones leaves PS=±RQ or PQ=±RS. (It’s laborious and nit-picking)

If PS=RQ, PR and PS are adjacents sides and A=|PRXPS|
If PS=-RQ=QR, PQ and PS are adjacent sides and A=|PQXPS|

( This is really the same as what I did before with a correction.)

Exercise: Check OP and then redo with some arbitrary labels. It works out.

9. ## Re: vectors - prove shape is a parallelogram

Originally Posted by Hartlw
1) Find PQ,PR,PS: two of these have to be adjacent sides of a quadrilateral.
2) Find QR,QS: One of these has to be a side other than the two adjacent sides of 1)
Actually, there is no correction. This is the same result as in my previous post, because if you eliminate obvious wrong pairings, you get:

PS=+/-QR or PR=+/-QS

If you don't like the wording of this, use wording in previous post. They say the same thing.

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