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There is a parallelogram that has the vertices 0, a, b, and a+b, all of which are three dimensional vectors.

a = \begin{pmatrix} 2 \\ -6 \\ 5 \end{pmatrix}b = \begin{pmatrix} -1 \\ -2 \\ 0 \end{pmatrix}

I know that the formula for two dimensional vectors is:

If u= $\begin{pmatrix} a \\ b \end{pmatrix}$ and v =$\begin{pmatrix} x \\ y \end{pmatrix}$, the parallelogram that has the vertices 0, a, b, and a+b has area $|ay-bx|$. What would the formula look like for three dimensional vectors?

Math is Life
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  • Have a look at Prof. Josiah Willard Gibbs' geometric definition of the cross (or skew) product, and realize that $A\sin \theta$ is just the height of a parallelogram. http://math.stackexchange.com/questions/420006/why-does-cross-product-give-a-vector-which-is-perpendicular-to-a-plane – John Joy Apr 27 '15 at 12:48

3 Answers3

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Form a matrix $A$ whose columns are the given vectors. Then $\det (A^TA)$ is the square of the area of the parallelogram. This generalizes easily to $k$ vectors in ${\Bbb R}^n$.

Tad
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A parallelogram with side lengths $a$ and $b$, with angle $\theta$ between them always has area

$$ab\sin\theta,$$

regardless of the dimension in which the parallelogram lives. Here, $a = \lVert {\bf a} \rVert$, $b$ is defined analogously, and you can find $\theta$ using the standard technique involving the dot product.

As you're seeing, there are many equivalent formulas for the area.

pjs36
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Take the cross product. Or, more generally for vectors in $\Bbb R^n$, take the wedge product.

http://en.wikipedia.org/wiki/Exterior_algebra

Jamie Lannister
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