The Area of a Parallelogram

Section 1.4 The Area of a Parallelogram

The purpose of this section is to derive a formula for the area of a parallelogram spanned by two vectors in the plane, in space or more generally in \(\mathbb R^N\text{.}\) The result will be essential for the understanding of the transformation formula for double integrals and the definition of surface integrals!

The area of a parallelogram is given by the product of its base and its height. If we look at the parallelogram spanned by two vectors \(\vect x\) and \(\vect y\) then the base is \(\|\vect x\|\text{,}\) and the height is \(\|\vect n\|\text{,}\) where \(\vect n\) is the projection of the vector \(\vect y\) into the direction orthogonal to \(\vect x\) as shown in Figure 1.15.

Figure 1.15. The Parallelogram spanned by \(\vect x\) and \(\vect y\text{.}\)

To compute \(\|\vect n\|\) note that

\begin{align*} \|\vect n\|^2\amp =\|\vect y-\vect p\|^2\\ \amp =\vect y\cdot\vect y -2\frac{\vect x\cdot\vect y}{\|\vect x\|^2}\vect x\cdot\vect y +\frac{(\vect x\cdot\vect y)^2}{\|\vect x\|^4}\vect x\cdot\vect x\\ \amp =\|\vect y\|^2 -2\frac{(\vect x\cdot\vect y)^2}{\|\vect x\|^2} +\frac{(\vect x\cdot\vect y)^2}{\|\vect x\|^4}\|\vect x\|^2\\ \amp =\|\vect y\|^2-\frac{(\vect x\cdot\vect y)^2}{\|\vect x\|^2}\text{.} \end{align*}

(This is the same computation as done in the proof of the Cauchy-Schwarz inequality).

Hence the area, \(A\text{,}\) of the parallelogram under consideration is

\begin{align} A\amp=\|\vect x\|\|\vect n\|\notag\\ \amp=\|\vect x\|\sqrt{\|\vect y\|^2 -\frac{(\vect x\cdot\vect y)^2}{\|\vect x\|^2}}\notag\\ \amp=\sqrt{\|\vect x\|^2\|\vect y\|^2-(\vect x\cdot\vect y)^2}\text{.}\tag{1.6} \end{align}

By definition of the determinant of a \(2\times 2\)-matrix and the norm of a vector we see that

\begin{align*} A^2\amp=\det \begin{bmatrix} \|\vect x\|^2\amp \vect x\cdot\vect y\\ \vect x\cdot\vect y\amp \|\vect y\|^2 \end{bmatrix}\\ \amp=\det \begin{bmatrix} \vect x\cdot\vect x\amp \vect x\cdot\vect y\\ \vect x\cdot\vect y\amp \vect y\cdot\vect y \end{bmatrix}\text{.} \end{align*}

Now let \(J:=\begin{bmatrix}\vect x\amp \vect y\end{bmatrix}\) denote the matrix with columns \(\vect x\) and \(\vect y\text{,}\) and let \(J^T\) be its transposed matrix. Taking into account Note 1.6 we find that

\begin{equation*} \begin{bmatrix} \vect x^T\vect x\amp \vect x^T\vect y\\ \vect x^T\vect y\amp \vect y^T\vect y \end{bmatrix} =J^TJ\text{.} \end{equation*}

Hence we have proved the following theorem.

Theorem 1.16. Area of parallelogram.

Suppose that \(\vect x,\vect y\in\mathbb R^N\text{.}\) Then the area of the parallelogram spanned by \(\vect x\) and \(\vect y\) is

\begin{equation} A=\sqrt{\det(J^TJ)}\text{,}\tag{1.7} \end{equation}

where \(J:=\begin{bmatrix}\vect x\amp \vect y\end{bmatrix}\) is the \(N\times 2\)-matrix with columns \(\vect x\) and \(\vect y\text{.}\) If \(N=2\text{,}\) that is, for vectors in the plane we have

\begin{equation*} A=|\det(J)|\text{.} \end{equation*}

Proof.

See the above discussion. The last assertion follows as \(\det(J^TJ)=\det J^T\det J=(\det J)^2\text{.}\)

Remark 1.17.

Note that in the formula (1.7) it is essential to compute \(J^TJ\) and not \(JJ^T\text{.}\) The matrix \(J^TJ\) is always symmetric, so only three entries need to be computed.

In a first example we compute the area of a parallelogram spanned by two vectors in the plane.

Example 1.18.

Compute the area of the parallelogram spanned by the vectors \((1,3)\) and \((2,4)\text{.}\)

Answer.

\(2\) square units.

Solution.

According to Theorem 1.16 the area is given by

\begin{equation*} \Biggl|\det \begin{bmatrix} 1\amp 2\\3\amp 4 \end{bmatrix}\Biggr| =|4-6|=2\text{ square units.} \end{equation*}

In a second example we compute the area of a parallelogram spanned by two vectors in three dimensional space.

Example 1.19.

Compute the area of the parallelogram spanned by \((1,2,4)\) and \((1,1,3)\text{.}\)

Answer.

\(\sqrt{6}\) square units.

Solution.

We form the matrix \(J\) with the given vectors as its columns and its transposed matrix:

\begin{equation*} J= \begin{bmatrix} 1\amp 1\\2\amp 1\\4\amp 3 \end{bmatrix} \qquad J^T= \begin{bmatrix} 1\amp 2\amp 4\\1\amp 1\amp 3 \end{bmatrix}\text{.} \end{equation*}

Then we form their product:

\begin{equation*} J^TJ= \begin{bmatrix} 1\amp 2\amp 4\\1\amp 1\amp 3 \end{bmatrix} \begin{bmatrix} 1\amp 1\\2\amp 1\\4\amp 3 \end{bmatrix} = \begin{bmatrix} 21\amp 15\\15\amp 11 \end{bmatrix}\text{.} \end{equation*}

Finally we take the determinant of the last matrix:

\begin{equation*} \det \begin{bmatrix} 21\amp 15\\15\amp 11 \end{bmatrix} =231-225=6\text{.} \end{equation*}

According to Theorem 1.16 the area of the parallelogram is

\begin{equation*} \sqrt{\det(J^TJ)}=\sqrt 6\text{ square units.} \end{equation*}

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