# Find area of parallelogram given vertices

1. Apr 25, 2014

### Zashmar

1. The problem statement, all variables and given/known data
Find the are of the parallelogram ABCD where A is (1,2,-3), B is (-1,3,-4) and D is (1,5,-2)

2. Relevant equations
Area=$\left|AxB\right|$
where A and B are the vectors AD, and AB respectively.

3. The attempt at a solution
I have calculated AD to be= (0,-3,-1)
and AB=(2,-1,1)

∴ to find $\left|AxB\right|$ I subbed the two vectors into a 3x3 matrix. Finding the magnitude of th result, I calculated the area to be 2x$\sqrt{11}$. The Anser is 2x$\sqrt{14}$. What have I done wrong?

Last edited: Apr 25, 2014
2. Apr 25, 2014

### SammyS

Staff Emeritus
You likely made an error in finding the vector product $\vec{A}\times \vec{B}$.

BTW: Use " \times " for $\times\$ .

3. Apr 25, 2014

### Zashmar

Well I will show you how i did it because I cannot find an error:

i j k
0-3 1
2-1 1
^that is a matrix

So i got:
-2i+2j+6k

=Sqrt 44
=2 sqrt 11

4. Apr 25, 2014

### jbunniii

Shouldn't the second row of your matrix be $(0, -3, -1)$ instead of $(0, -3, 1)$?

5. Apr 25, 2014

### Staff: Mentor

The first vector is actually DA, and the second is BA.
AD is -(DA) = <0, 3, 1>
and AB = -(BA) = <-2, 1, -1>.

6. Apr 25, 2014

### SammyS

Staff Emeritus
The vectors you have in the OP will work. (They both have the sign error Mark pointed out.) .. but they will give $2 \, \sqrt{14} \$ .