I have this one Linear Algebra question that is asking me to compute the area of a parallelogram defined by 4 vectors. Here is the question:
Let $\vec{u}=\begin{bmatrix}a\\b\end{bmatrix}$ and
$\vec{v}=\begin{bmatrix}c\\0\end{bmatrix}$, where $a$, $b$, and $c$
are positive. Compute the area of the parallelogram determined by
$\vec{u}$, $\vec{v}$, $\vec{u}+\vec{v}$, and $\vec{0}$, and compute
the determinants of the matrices $\begin{bmatrix}\vec{u} &
\vec{v}\end{bmatrix}$ and $\begin{bmatrix}\vec{v} &
\vec{u}\end{bmatrix}$. Draw a picture and explain what you find.
Here is what I have so far:
$\begin{bmatrix}\vec{u} & \vec{v}\end{bmatrix}=\begin{bmatrix}a & c\\ b & 0\end{bmatrix} \rightarrow \begin{vmatrix}a & c\\ b & 0\end{vmatrix}=-bc$
$\begin{bmatrix}\vec{v} & \vec{u}\end{bmatrix}=\begin{bmatrix}c & a\\ 0 & b\end{bmatrix} \rightarrow \begin{vmatrix}c & a\\ 0 & b\end{vmatrix}=bc$
I also know that $\vec{u}+\vec{v}=\begin{bmatrix}a\\b\end{bmatrix}+\begin{bmatrix}c\\0\end{bmatrix}=\begin{bmatrix} a+c\\b \end{bmatrix}$
The part that I am confused at is finding the area of the parallelogram… How would I go about doing that with 4 vectors? Thanks.
Best Answer
To elaborate: The height is $|u|sin(\theta)$ and the height is $|v|$. Then you can find $\theta$ as $u \cdot v = |u||v|cos(\theta)$. So the area should be $A=|u||v|sin(cos^{-1}(\frac{u \cdot v}{|u||v|}))$
Also, the algebraic answer is equivalent. $|u \times v|$ would yield the same result which you calculated already.