Linear Algebra Help, Area of a Parallelogram using vectors

[Axoren]

Homework Statement

Find the area of the parallelogram defined by the vectors

v = {1 1 3 1}
w = {-2 -1 2 2}

Homework Equations

Area = v dot w * sin(theta)
theta = cos^-1(v dot w / |v|*|w|)

The Attempt at a Solution

Solved
General Solution:

Area of a parallelogram for non-R^3 vectors = v dot w * sin(cos^-1(v dot w / (|v|*|w|)))

 

[jegues]

If v and w are two vectors representing two adjacent sides of the parallelogram, then the area is the magnitude of the cross product of those two vectors.

 

[Axoren]

If v and w are two vectors representing two adjacent sides of the parallelogram, then the area is the magnitude of the cross product of those two vectors.

You can't perform cross product on vectors outside of R^3

 

[jegues]

You can't perform cross product on vectors outside of R^3

Whoops! Sorry, I didn't notice they were 4 dimensional vectors.

 

[Axoren]

I got it, some how. Updated original post.

 

[jegues]

I got it, some how. Updated original post.

What do you mean you got it some how... What did you do? Alegbra mistake?

 

[Axoren]

What do you mean you got it some how... What did you do? Alegbra mistake?

I didn't depend on the cross product and started using other equations around the internet.

I provided them in the original post.

 

[hgfalling]

I don't think that's right. You can let

$$A=\begin{bmatrix} 1 & -2 \\ 1 & -1 \\ 3 & 2 \\ 1 & 2 \end{bmatrix}$$

and then the volume is:

$$V = \sqrt{det \left(A^TA \right)}$$

which isn't 5.

EDIT: Oh, I see you got it with some other formula. Was the answer $\sqrt{131}$ ?