Linear Algebra: Compute Area of Parallelogram

Question

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.

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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.

Answer 1

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.

Answer 2

Try to break the area down into two triangles, each triangle is the area between to vectors where the starting points touch. Put the two groups of two vectors into a matrix as columns and find the determinant/2 for each group, the sum should be the area.