# Vectors - area of parallelogram

• March 17th 2011, 12:31 PM
sderosa518
Vectors - area of parallelogram
Find an area of a parallelogram determined by the points P(7,-5,5), Q(-5,10,10), R(-8,8,-4) and S(-20,23,1).

-->
PQ = -12i + 15j + 5k
-->
RS = -12i + 15j + 5k
--
>
PS = -27i + 28j - 4k
-->
RQ = 3i -2j -14k

* This is what I have, I am lost at the moment. I have PQ and RS are parallel. I know I need to use the cross product. Any assistance on what to do next????
• March 17th 2011, 12:38 PM
Ackbeet
I would actually compute PR and SQ. You want to use the cross product of two non-parallel outer sides (the area is the magnitude of said cross product). Because PS and RQ are not parallel or anti-parallel, you know that they are diagonals, and hence not so useful (you might be able to use them, but I'm not entirely sure how).
• March 17th 2011, 12:48 PM
Plato
The area is $\left\| {\overrightarrow {PQ} \times \overrightarrow {PS} } \right\|$

Is your question why is that the area?
• March 17th 2011, 01:01 PM
sderosa518
The objective is to find the area of the parallelogram determined by the folowing points:

P(7,-5,5), Q(-5,10,10), R(-8,8,-4) and S(-20,23,1)

Basically I just need to solve the following:

The area is PQ x PS????
• March 17th 2011, 01:14 PM
Plato
Quote:

Originally Posted by sderosa518
The objective is to find the area of the parallelogram determined by the folowing points:
P(7,-5,5), Q(-5,10,10), R(-8,8,-4) and S(-20,23,1)
The area is PQ x PS????

NO It is $\|\vec{PQ}\times\vec{PS}\|$.

Suppose that $\vec{a}~\&~\vec{b}$ are adjacent sides of a parallelogram.

Then the area of that parallelogram is $\|\vec{a}\|\|\vec{b}\|\sin(\theta)$, where $\theta$ is the angle between the vectors.

Recall that $\sin (\theta ) = \dfrac{{\left\| {\overrightarrow a \times \overrightarrow b } \right\|}}{{\left\| {\overrightarrow a } \right\|\left\| {\overrightarrow b } \right\|}}$.
• March 17th 2011, 01:44 PM
sderosa518
this is wat i got:

-->
PQ = -12i + 15j + 5k
-->
PS = -27i + 28j - 4k

i j k
PQ x PS = -12 15 5
-27 28 -4

= -200i + 357j + 69k

= ll -200^2 + 357^2 + 69^2 ll

= ll172210ll
= 1312.253
• March 17th 2011, 02:05 PM
Plato
Quote:

Originally Posted by sderosa518
-->
PQ = -12i + 15j + 5k CORRECT
-->
PS = -27i + 28j - 4k CORRECT
i j k
PQ x PS = -12 15 5
-27 28 -4
= -200i + 357j + 69k WRONG

$\vec{PQ}\times\vec{PS}=-200i-183j+69k$
• March 17th 2011, 02:25 PM
sderosa518
It doesnt seem right. May I ask why you choose PQ and PS???
• March 17th 2011, 02:43 PM
Plato
Quote:

Originally Posted by sderosa518
It doesnt seem right. May I ask why you choose PQ and PS???

Those two vectors are adjacent sides of a parallelogram.

You just made an arithmetic error in the matrix of the cross product.
• March 17th 2011, 02:58 PM
sderosa518