Area of a Parallelogram by Geometric Algebra

Abstract: We show here how to relate the area of a parallelogram in the usual manner to how it can be represented in geometric algebra.

Download PDF file: Area of a Parallelogram (by Geometric Algebra)

Our basic goal here is to represent the area of a parallelgram in geometric algebra, connecting it up with its standard presentation in geometry.

         
Figure 1. We have here a typical parallelogram.

In Fig. 1 we see a parallelogram whose area we know from high school math to be \begin{equation} \text{Area Parallelogram} = ab\sin\theta\,,\label{eq:absintheta} \end{equation} where $\theta$ is the angle between sides $\ba$ and $\bb$, and $a$ and $b$ are, respectively, the lengths of sides $\ba$ and $\bb$.

Let's now place vectors on the sides of the parallelogram so that we can represent its area with a bivector (which better than a cross product).

         

Figure 2. Written in bivector form, the area of
the parallelogram can be expressed as $\ba\wedge\bb$
.

Now, let $\bB$ be any nonzero bivector represented as the wedge product of any two vectors (such as in Fig.\ 2). In geometric algebra, the magnitude of this bivector is defined by \begin{equation} \abs{\bB} = [\langle \bB^\dagger \bB \rangle]^{1/2}\,, \end{equation} where $\langle\cdots \rangle = \langle\cdots \rangle_0$ means to take the scalar part of what's between the brackets. The symbol ${}^\dagger$ is the reverse operator, which means to take the ordering of the vectors in reverse order. This operation distributes over addition. Scalars and vectors are invariant under the reverse operation. For clarification, with the $a_i$'s as vectors, we have the general relations: \begin{align} (a_1a_2\cdots a_n)^\dagger &= a_n\cdots a_2a_1\,,\notag\\ (a_1\wedge a_2\wedge\cdots \wedge a_n)^\dagger &= a_n\wedge \cdots\wedge a_2\wedge a_1\notag\\ (a_1\wedge a_2)^\dagger &= - a_1\wedge a_2\,. \end{align}

Some additional background in geometric algebra may be of help. The geometric product of two vectors $\ba$ and $\bb$ is given as \begin{equation} \ba\bb = \ba\cdot\bb +\ba\wedge\bb\,. \end{equation} Solving for the bivector part, we have \begin{equation} \ba\wedge\bb = \ba\bb - \ba\cdot\bb \,. \end{equation} We also need to know that \begin{equation} \ba\cdot\bb = \half(\ba\bb + \bb\ba) \,, \end{equation} and, of course, the familiar relation from vector algebra, \begin{equation} \ba\cdot\bb = ab\cos\,\theta \,. \end{equation}

Now, for any two multivectors $A$ and $B$ $$\begin{equation}\label{eq:<AB>dagger} \langle A B \rangle^\dagger = \langle B^\dagger A^\dagger \rangle\,. \end{equation}$$ As a corollary, if $A$ and $B$ are both bivectors, then \begin{equation} \langle A B \rangle^\dagger = \langle (-B)(- A) \rangle = \langle B A \rangle\,. \end{equation} My first task is to show that $ \abs{\bB}$ is the area of the bivector parallelogram, and is given as $ab\sin\theta$. For convenience, I'll start with $ \abs{\bB}^2$ where $\bB = \ba\wedge\bb$. \begin{align} \abs{\bB}^2 &= \langle \bB^\dagger \bB \rangle\notag\\ &= \langle ( \ba\wedge\bb )^\dagger \ba\wedge\bb \rangle\notag\\ &= \langle ( \ba\bb - \ba\cdot\bb )^\dagger ( \ba\bb - \ba\cdot\bb )\rangle\notag\\ &= \langle ( \bb\ba - \ba\cdot\bb )( \ba\bb - \ba\cdot\bb )\rangle\notag\\ &= \langle \bb\ba\ba\bb - \ba\cdot\bb (\bb\ba+ \ba\bb) + ( \ba\cdot\bb )^2\rangle\notag\\ &= \langle \ba^2\bb^2 - 2(\ba\cdot\bb)^2 + ( \ba\cdot\bb )^2\rangle\notag\\ &= a^2b^2 - a^2b^2 \cos^2\theta\notag\\ &= a^2b^2\sin^2\theta\,. \end{align} On taking square roots of both sides, this result agrees with what we knew from (\ref{eq:absintheta}). Now, a useful corollary: if $A$ is a scalar or a vector, then \begin{equation}\label{eq:A_dagger=A} A^\dagger = A\,. \end{equation} My last task is to show that \begin{equation} \abs{\bB} = |\, \bB\cdot\bsigma_1\wedge\bsigma_2\,|\,,\label{eq:abs(B)} \end{equation} where $\bsigma_1$ and $\bsigma_2$ are unit vectors along the $x$- and $y$-axes, respectively. Next follows some useful prerequisite results: Let $A$ and $B$ be any two multivectors. If $\bh$ is any multivector such that $\bh\bh^\dagger = 1$, then \begin{equation} A B = A(1)B = A\bh\bh^\dagger B = (A\bh)(\bh^\dagger B)\,, \end{equation} which comes from the fact that the geometric product is associative! (Yay!) In particular, let $\bh = \bsigma_1\bsigma_2$, then $\bh\bh^\dagger = 1$. (Prove it! Hint: First, $\bsigma_1^2 = \bsigma_2^2 = 1$, and, second, $\bh\bh^\dagger=\bsigma_1\bsigma_2\bsigma_2\bsigma_1$, and then employ associativity carefully and remember that scalars commute with any multivector.) So, \begin{equation} A B = A(1)B = (A\bsigma_1\bsigma_2)(\bsigma_2\bsigma_1B)\,. \end{equation} Now, $$\begin{align} \abs{\bB}^2 &= \langle \bB^\dagger \bB \rangle\notag\\ &= \langle \bB^\dagger\bsigma_1\bsigma_2\bsigma_2\bsigma_1 \bB \rangle\notag\\ &= \langle (\bB^\dagger\bsigma_1\bsigma_2)(\bsigma_2\bsigma_1 \bB) \rangle\notag\\ &= \langle (\bB^\dagger\cdot\bsigma_1\wedge\bsigma_2)(\bsigma_2\wedge\bsigma_1\cdot \bB) \rangle\notag\\ &= (\bB^\dagger\cdot\bsigma_1\wedge\bsigma_2)\langle(\bsigma_2\wedge\bsigma_1\cdot \bB) \rangle\notag\\ &= (\bB^\dagger\cdot\bsigma_1\wedge\bsigma_2)\langle(\bsigma_2\wedge\bsigma_1\cdot \bB) \rangle^\dagger \quad\text{from (\ref{eq:A_dagger=A})}\notag\\ &= (\bB^\dagger\cdot\bsigma_1\wedge\bsigma_2)^2 \quad\text{from (\ref{eq:<AB>dagger})}\notag\\ &= |\,\bB^\dagger\cdot\bsigma_1\wedge\bsigma_2\,|^2\notag\\ &= |\,\bB\cdot\bsigma_1\wedge\bsigma_2\,|^2\,. \end{align}$$ We get (\ref{eq:abs(B)}) by taking square roots of both sides. QED

Additional Note. The reader who is not familiar with geometric algebra and wedge products, can reference on-line sources through the search string 'wedge product'. The main reference book is New Foundations for Classical Mechanics by David Hestenes (Kluwer Academic Publishers). An online reference is

Download PDF file: A Geometric Algebra Primer by David Hestenes