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?
Best Answer
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 ${\Bbb R}^n$.