Question

The vectors $\mathbf{A}=a\mathbf{i}+b\mathbf{j}$and $\mathbf{B}=c\mathbf{i}+d\mathbf{j}$form two sides of a parallelogram. Show that the area of the parallelogram is given by the absolute value of the following determinant.

role="math" localid="1664258424007" $\left|\begin{array}{cc}\text{a}& \text{b}\\ \text{c}& \text{d}\end{array}\right|$

Short answer

The area of a parallelogram whose nonparallel sides are represented by vectors$\text{A}=ai+bj$ and $\text{B}=c\text{i}+dj$is equal to$\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|$ .

Step-by-step solution

Given information.

The nonparallel sides of given parallelogram are represented by vectors $\text{A}=ai+bj$and$\text{B}=c\text{i}+dj$ .

Area of a triangle.

The area of the triangle isgiven by,

$\mathbf{▵}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\times }\mathit{b}\mathbf{\times }\mathit{h}$

Find the area of a parallelogram.

Let PQRS be the given parallelogram with side PS and PQ are represented by vectors A and B respectively as shown in figure.

Let the vector A and B represents the sides PS and PQ respectively. Then $\overrightarrow{\text{PS}}=\overrightarrow{\text{A}}\Rightarrow \text{PS}=\left|\overrightarrow{\text{A}}\right|=\text{QR}$, and $\overrightarrow{\text{PQ}}=\overrightarrow{\text{B}}\Rightarrow \text{PQ}=\left|\overrightarrow{\text{B}}\right|$. Let $\text{QM}=\text{SN}=h$be the heights of triangles $△\text{PQR}$ and $△\text{QSR}$ made by the diagonal $\text{QS}$. Further, let $\angle \text{QPR}=\theta$, then height $h=\text{PQ}\sin \theta$.

Area of parallelogram ,

$\text{PQRS}=▵\text{PQS}+▵\text{QRS}$

But

$\begin{array}{rcl}▵\text{PQS}& =& \frac{1}{2}·\text{PS}·h\\ & =& \frac{1}{2}·\text{PS}·\text{PQ}\sin \theta \\ & =& \frac{1}{2}·\text{PS}·\text{PQ}\sin \theta \\ & =& \left|\overrightarrow{\text{A}}\right|·\left|\overrightarrow{\text{B}}\right|\sin \theta \because \left|\overrightarrow{\text{PQ}}\right|=\left|\overrightarrow{\text{B}}\right|\end{array}$

Solve further

$\begin{array}{rcl}▵\text{QRS}& =& \frac{1}{2}·\text{QR}·h\\ & =& \frac{1}{2}·\text{PS}·\text{PQ}\sin \theta \\ & =& \frac{1}{2}·\text{PS}·\text{PQ}\sin \theta \\ & =& |\overrightarrow{\text{A}\mid }·|\overrightarrow{\text{B}}\mid \sin \theta \end{array}$

Therefore, QR and PS are parallel and equal sides of parallelogram, and $\text{QM}=h=\text{SN}$the distances between two parallel sides.

Therefore,

$\begin{array}{rcl}\text{PQRS}& =& \frac{1}{2}·|\overrightarrow{\text{A}\mid }·|\overrightarrow{\text{B}}\left|\sin \theta \hat{k}+\frac{1}{2}·\right|\overrightarrow{\text{A}\mid }·\left|\overrightarrow{\text{B}}\right|\sin \theta \hat{k}\\ & =& \left|\overrightarrow{\text{A}}\right|·\left|\overrightarrow{\text{B}}\right|\sin \theta \hat{k}=(\overrightarrow{\text{A}}\times \overrightarrow{\text{B}})\hat{k}\end{array}$

Find the absolute value of a determinant

Now, by the definition of vector product,

$\begin{array}{rcl}\text{A}\times \text{B}& =& \left(a\overrightarrow{i}+b\overrightarrow{j}\right)\times \left(c\overrightarrow{i}+d\overrightarrow{j}\right)\\ & =& (a·d)(\overrightarrow{i}\times \overrightarrow{j})+(b·c)(\overrightarrow{j}\times \overrightarrow{i})\end{array}$

(localid="1664261784809" $(\overrightarrow{i}\times \overrightarrow{j})=\hat{k}$unit vector perpendicular to the plane PQRS,localid="1664261796324" $(\overrightarrow{i}\times \overrightarrow{i})=(\overrightarrow{j}\times \overrightarrow{j})=0$, and $(\overrightarrow{j}\times \overrightarrow{i})=-\hat{k}$)

Therefore,

localid="1664261396291" $\begin{array}{rcl}d(\overrightarrow{\text{A}}\times \overrightarrow{\text{B}})& =& (a·d)\left(\hat{k}\right)+(b·c)(-\hat{k})\\ & =& (a·d-b·c)\left(\hat{k}\right)\end{array}$

localid="1664261571379" $\left|\begin{array}{cc}\overrightarrow{A}\times & \overrightarrow{B}\end{array}\right|=\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|$

Thus, the area of a parallelogram whose nonparallel sides are represented by vectors and is equal to localid="1664261602075" $\left|\begin{array}{cc}\overrightarrow{A}\times & \overrightarrow{B}\end{array}\right|=\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|$.

Hence, proved.