Finding the Area of a Parallelogram with Two Vectors in R^4

[ianmc7]

I need to find the area of a parallelogram with two vectors in R^4 my book has nothing on this and I don't know how to do it.

 

[rock.freak667]

If the area of a parallelogram is $absin\theta$, and a,b are vectors. Then you can re-write the area as:

$$|\vec{a}|| \vec{b}| sin\theta$$


Does that look familiar?

 

[dynamicsolo]

If you're proposing the use of the vector cross product, rock.freak667, be aware that the definition is somewhat problematic in $$R^n$$ for n > 3 (for instance, one source claims it only exists in R^3 and R^7: http://everything2.com/title/Vector%20product , though there are other remarks on this around the 'Net).

Since the two vectors would still span a plane in higher-dimensional space, the definition of area for the parallelogram produced by the vectors would still be meaningful. You could use the vector dot product in the usual way to find the cosine of the angle (and thus the angle) between the vectors in that plane (since the methods for calculating vector length and vector dot product, happily, do generalize to n dimensions). It will still be true that

$$A = |\vec{a}|| \vec{b}| sin\theta$$

 

[pkleinod]

Since the two vectors would still span a plane in higher-dimensional space, the definition of area for the parallelogram produced by the vectors would still be meaningful...

$$A = |\vec{a}|| \vec{b}| sin\theta$$

Yes, that is fine, but here is
another, more direct, way to get the area by calculating the magnitude
of the bivector representing the parallelogram. In terms of a set of
orthonormal basis vectors $$\{ e_i }, i = 1,N \}$$ in N dimensions. The vectors are
$$\vec{a} = \sum_i a_i e_i$$
$$\vec{b} = \sum_i b_i e_i$$

The parallelogram is represented by the outer product or these two vectors:

$$\vec{a}\wedge \vec{b} = \sum_{i<j} (a_ib_j - a_jb_i)e_ie_j$$

The square of the area is then

$$|\vec{a}\wedge\vec{b}|^2 = \sum_{i<j}(a_i b_j - a_j b_i)^2.$$

Using this approach, it is not necessary to compute the angle $$\theta$$.
Note that $$a\wedge b$$ is NOT the cross product and exists for all dimensions.

 

[dynamicsolo]

Thank you for the information on the "wedge product", an operation which has yet to penetrate far enough into the basic curriculum. (I am not familiar enough with it myself as yet, so I chose a method using devices that would be familiar in typical first courses in vector operations.)

"Ah, wedge product... is there nothing you can't do...?"

 

[pkleinod]

"Ah, wedge product... is there nothing you can't do...?"

Ah! That's the spirit! Does the wedge product ever fail?
' "No, never."
"What, never?"
"Well, hardly ever!" ' (G&S: HMS Pinafore)

To tickle your curiosity: In the 5-dimensional conformal model of Euclidean 3D space, the expression for a sphere in terms of 4 vectors (in the 5D space) to points on the surface of the sphere is

$$p\wedge q\wedge r\wedge s$$

(I am not familiar enough with it myself as yet...)

You can change this by going to this site: http://www.geometricalgebra.net/
And if you want to have FUN at the same time, click on the "downloads" link and get the GAViewer and the list of interactive figures. Enjoy!