Find area of parallelogram given vertices

[Zashmar]

Homework Statement

Find the are of the parallelogram ABCD where A is (1,2,-3), B is (-1,3,-4) and D is (1,5,-2)

Homework Equations

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

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?

 

[SammyS]

Homework Statement

Find the are of the parallelogram ABCD where A is (1,2,-3), B is (-1,3,-4) and D is (1,5,-2)

Homework Equations

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

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?

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

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

 

[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

 

[jbunniii]

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

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

 

[Mark44]

Homework Statement

Find the are of the parallelogram ABCD where A is (1,2,-3), B is (-1,3,-4) and D is (1,5,-2)

Homework Equations

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

The Attempt at a Solution

I have calculated AD to be= (0,-3,-1)
and AB=(2,-1,1)

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

 

[SammyS]

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

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} \$ .