The Area of a Parallelogram in 2-Space

# The Area of a Parallelogram in 2-Space

Recall that if we have two vectors $\vec{u}, \vec{v} \in \mathbb{R}^3$, the area of the parallelogram defined by then can be calculated with the formula $A = \| \vec{u} \times \vec{v} \| = \| \vec{u} \| \| \vec{v} \| \sin \theta$. We will now look at a formula for calculating a parallelogram of two vectors in $\mathbb{R}^2$.

 Theorem 1: If $\vec{u}, \vec{v} \in \mathbb{R}^2$ then the area of the parallelogram formed by $\vec{u}, \vec{v}$ is $A = \mathrm{abs} \begin{vmatrix} u_1 & u_2\\ v_1 & v_2 \end{vmatrix}$.
• Proof: Since the cross product is defined only in 3-space, we will derive the following formula to calculate the area of a parallelogram in 2-space by taking our vectors $\vec{u} = (u_1, u_2)$ and $\vec{v} = (v_1, v_2)$ and placing them in $\mathbb{R}^3$, that is letting $\vec{u} = (u_1, u_2, 0)$ and $\vec{v} = (v_1, v_2, 0)$.
• First create a $3 \times 3$ matrix with the first row entries being the standard unit vectors $\vec{i}$, $\vec{j}$, and $\vec{k}$. The second row will be the components of $\vec{u}$ (with the last being 0) and the third row will be the components of $\vec{v}$ (with the last being 0). We will then take the determinant of this matrix:
(1)
\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ u_1 & u_2 & 0\\ v_1 & v_2 & 0 \end{vmatrix}
• If we evaluate this determinant by cofactor expansion along the first row, we obtain that:
(2)
\begin{align} \vec{u} \times \vec{v} = \vec{i} \begin{vmatrix} u_2 & 0 \\ v_2 & 0 \end{vmatrix} - \vec{j} \begin{vmatrix} u_1 & 0 \\ v_1 & 0 \end{vmatrix} + \vec{k} \begin{vmatrix} u_1 & u_2 \\ v_1 & v_2 \end{vmatrix} \end{align}
• However, we note that the determinants $\vec{i} \begin{vmatrix} u_2 & 0 \\ v_2 & 0 \end{vmatrix} = 0$ and $- \vec{j} \begin{vmatrix} u_1 & 0 \\ v_1 & 0 \end{vmatrix} = 0$. Thus it follows that:
(3)
\begin{align} \vec{u} \times \vec{v}= \vec{k} \begin{vmatrix} u_1 & u_2 \\ v_1 & v_2 \end{vmatrix} \end{align}
• Now let's take the norm of both sides to obtain that:
(4)
\begin{align} \| \vec{u} \times \vec{v} \| = \| \vec{k} \| \mathrm{abs} \begin{vmatrix} u_1 & u_2 \\ v_1 & v_2 \end{vmatrix} \end{align}
• However, $\vec{k}$ is a standard unit vector so $\| \vec{k} \| = 1$ and thus $\| \vec{u} \times \vec{v} \| = \mathrm{abs} \begin{vmatrix}u_1 & u_2 \\ v_1 & v_2 \end{vmatrix}$, and of course, this is equal to the area of the parallelogram. $\blacksquare$