# Math Help - vectors - prove shape is a parallelogram

1. ## vectors - prove shape is a parallelogram

I know I have to show that two sides are equal and in order to prove this is a parallelogram, however I am not sure which sides to pick ?

i.e.

Is it PQ = RS

or

RP = SQ,

How would I know, and I dont think i would be able to sketch the diagram as its in a R^{3}

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

use a graph paper and form the quadrilateral defined by the given vectors...the rest is easy...

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

Originally Posted by MINOANMAN
use a graph paper and form the quadrilateral defined by the given vectors...the rest is easy...
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.

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

Originally Posted by Tweety
I know I have to show that two sides are equal and in order to prove this is a parallelogram, however I am not sure which sides to pick ?

i.e.

Is it PQ = RS

or

RP = SQ,

How would I know, and I dont think i would be able to sketch the diagram as its in a R^{3}
What is a parallelogram? A shape with two sets of parallel sides. So set up the four vectors, show that two pairs of them are parallel...

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

so its guessing game?

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

no.

First make a rough sketch so you can tell which points are which vertices.

Then you can determine the vector of each of the four sides as the differences of the position vectors of the appropriate vertices.

To show the opposite sides are parallel you can show their cross products are zero.

Then use the cross product on the vectors of two adjacent sides to find the area.

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

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.

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

Originally Posted by Hartlw
Without any preconceptions 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)
You are making several preconceptions. You are assuming the four points are coplanar. You don't know that.

Without that your statement "PQ,PR,PS: two of these have to be adjacent sides of a quadrilateral" need not be true.

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

Hello, Tweety!

Given points: . $P(\text{-}1,0,2),\;Q(3,4,\text{-}1),\;R(3,2,\text{-}3),\;S(\text{-}1,\text{-}2,0)$
Show that $PQRS$ is a parallelgram and use the cross-product to find its area.

I assumed that PQRS are in order around the parallelogram.
Then I checked this assumption.

Code:
      (-1,0,2)   (3,4,-1)
P *---------* Q
/         /
/         /
/         /
S *---------* R
(-1,2,0)   (3,2,-3)
$\begin{array}{cccc}\text{We find that:} & \overrightarrow{PQ} \;=\;\langle 4,4,\text{-}3\rangle \\ & \overrightarrow{SR} \;=\;\langle 4,4,\text{-}3\rangle \end{array}$

$\begin{array}{ccc}\text{Hence: } & \overrightarrow{PQ}\;\;||\;\;\overrightarrow{SR} \\ & |\overrightarrow{PQ}| \:=\:|\overrightarrow{SR}| \end{array}$

If two sides of a quadrilateral are parallel and equal,
, , the quadrilateral is a parallelogram.

$\overrightarrow{PQ} \times \overrightarrow{QR} \;=\;\langle 4,4,\text{-}3\rangle \times \langle 0,\text{-}2,\text{-}2\rangle$

. . . . . . . . $=\; \begin{vmatrix}i&j&k \\ 4&4&\text{-}3 \\ 0&\text{-}2&\text{-}2 \end{vmatrix}$

. . . . . . . . $=\;i(\text{-}8-6) - j(\text{-}8-0) + k(\text{-}8-0)$

. . . . . . . . $=\;\text{-}14i + 8j - 8k$

$\text{Area} \:=\:\left|\overrightarrow{PQ} \times \overrightarrow{QR}\right| \:=\:\sqrt{(\text{-}14)^2 + 8^2 + (\text{-}8)^2} \:=\:\sqrt{324} \:=\:18$

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

Originally Posted by Plato
You are making several preconceptions. You are assuming the four points are coplanar. You don't know that.

Without that your statement "PQ,PR,PS: two of these have to be adjacent sides of a quadrilateral" need not be true.
I am not assuming coplanar. Two of PQ, PR, PS have to be adjacent sides of a quadrilateral.

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

Originally Posted by Hartlw
I am not assuming coplanar. Two of PQ, PR, PS have to be adjacent sides of a quadrilateral.
Well you clearly do not understand $\mathbb{R}^3$.

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

Originally Posted by Plato
Well you clearly do not understand $\mathbb{R}^3$.
What is the basis of that statement?

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

Originally Posted by Hartlw
What is the basis of that statement?
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.

14. ## 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.
he did say only two of them, not all three. I think we all can agree that two vectors are coplanar.

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

Originally Posted by romsek
he did say only two of them, not all three. I think we all can agree that two vectors are coplanar.
If you read my reply carefully, I did say two of. The point being that $P,~Q,~R,~S$ are the vertices of a quadrilateral only if the four points are coplanar.

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