# Linear Algebra: Compute Area of Parallelogram

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.

• The $4$ vectors form the corners of your parallelogram. In other words, this is the parallelogram you get by putting together $u$ and $v$. – Omnomnomnom Mar 10 '15 at 21:26
• In order to find the area of the parallelogram, draw it out and use the formula $$\text{Area} = \text{Base}\times\text{Height}$$ – Omnomnomnom Mar 10 '15 at 21:27
• @Omnomnomnom, so would the area be $-bc$? Since it is $\begin{bmatrix} \vec{u}&\vec{v} \end{bmatrix}$? – KFC Mar 10 '15 at 21:36
• Yes, and also because the base is $b$ and the height is $c$. – Omnomnomnom Mar 10 '15 at 21:37

## 1 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.

• So it would be the absolute value of the cross product? So, instead of $-bc$ it will be $bc$, right? – KFC Mar 10 '15 at 21:43
• Yes. It is the magnitude of the cross product. In the the general case, you square each component and then take the square root. For this case in which the cross product is just a number, it is just the absolute value. If you look at the formula $A=|u||v|sin(cos^{-1}(\frac{u \cdot v}{|u||v|}))$, the angle $\theta = cos^{-1}(\frac{u \cdot v}{|u||v|})$ has to be between $0$ and $\pi$ by definition of $cos^{-1}$. Thus that would make $sin(\theta) \geq 0$. Thus $A \geq 0$. – Jack Mar 10 '15 at 21:51