# Linear Algebra Help, Area of a Parallelogram using vectors

1. Aug 23, 2010

### Axoren

1. The problem statement, all variables and given/known data
Find the area of the parallelogram defined by the vectors

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

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

3. 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|)))

Last edited: Aug 23, 2010
2. Aug 23, 2010

### 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.

3. Aug 23, 2010

### Axoren

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

4. Aug 23, 2010

### jegues

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

5. Aug 23, 2010

### Axoren

I got it, some how. Updated original post.

6. Aug 23, 2010

### jegues

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

7. Aug 23, 2010

### Axoren

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

I provided them in the original post.

8. Aug 23, 2010

### 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 $\sqrt{131}$ ?