3

I want to calculate the area of a parallelogramm given the following four vertices:

$$\vec{p}=\begin{pmatrix}2 \\ 0\\3 \end {pmatrix},\vec{q}=\begin{pmatrix}8 \\ 1\\1 \end {pmatrix},\vec{r}=\begin{pmatrix}6 \\ -2\\-1 \end {pmatrix},\vec{s}=\begin{pmatrix}12 \\ -1\\-3 \end {pmatrix}$$

I know I need to find two sides of the parallelogramm and then take the magnitude of the cross product. Here is my problem:

Assume that I chose to compute $\vec{q}-\vec{p}$ and $\vec{s}-\vec{p}$. My picture could look like this:

enter image description here

However, since I don't know the spatial location of $p,q,r,s$ then the picture could also look like this:

enter image description here

So how do I determine which points to subtract in order to get the correct pair of vectors? Does it even matter which pair I take or will the area be the same because of the symmetry of the problem?

Nullspace
  • 979

1 Answers1

3

It doesn't matter. Either way gives you the same value.

For your second drawing, move the top half of the parallelogram to the right of the lower half. It's the same area.

marty cohen
  • 107,799