91Ó°ÊÓ

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

The area of the triangle isgiven by,

$\mathbf{â–\mu }\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{×}\mathit{b}\mathbf{×}\mathit{h}$

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 $\stackrel{→}{\text{PS}}=\stackrel{→}{\text{A}}⇒\text{PS}=\left|\stackrel{→}{\text{A}}\right|=\text{QR}$, and $\stackrel{→}{\text{PQ}}=\stackrel{→}{\text{B}}⇒\text{PQ}=\left|\stackrel{→}{\text{B}}\right|$. Let $\text{QM}=\text{SN}=h$be the heights of triangles $△\text{PQR}$ and $△\text{QSR}$ made by the diagonal $\text{QS}$. Further, let $∠\text{QPR}=\mathrm{θ}$, then height $h=\text{PQ}\sin \mathrm{θ}$.

Area of parallelogram ,

$\text{PQRS}=â–\mu \text{PQS}+â–\mu \text{QRS}$

But

$\begin{array}{rcl}â–\mu \text{PQS}& =& \frac{1}{2}·\text{PS}·h\\ & =& \frac{1}{2}·\text{PS}·\text{PQ}\sin \mathrm{θ}\\ & =& \frac{1}{2}·\text{PS}·\text{PQ}\sin \mathrm{θ}\\ & =& \left|\stackrel{→}{\text{A}}\right|·\left|\stackrel{→}{\text{B}}\right|\sin \mathrm{θ}âˆ\mu \left|\stackrel{→}{\text{PQ}}\right|=\left|\stackrel{→}{\text{B}}\right|\end{array}$

Solve further

$\begin{array}{rcl}â–\mu \text{QRS}& =& \frac{1}{2}·\text{QR}·h\\ & =& \frac{1}{2}·\text{PS}·\text{PQ}\sin \mathrm{θ}\\ & =& \frac{1}{2}·\text{PS}·\text{PQ}\sin \mathrm{θ}\\ & =& |\stackrel{→}{\text{A}âˆ\pounds }·|\stackrel{→}{\text{B}}âˆ\pounds \sin \mathrm{θ}\end{array}$

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

Therefore,

$\begin{array}{rcl}\text{PQRS}& =& \frac{1}{2}·|\stackrel{→}{\text{A}âˆ\pounds }·|\stackrel{→}{\text{B}}\left|\sin \mathrm{θ}\hat{k}+\frac{1}{2}·\right|\stackrel{→}{\text{A}âˆ\pounds }·\left|\stackrel{→}{\text{B}}\right|\sin \mathrm{θ}\hat{k}\\ & =& \left|\stackrel{→}{\text{A}}\right|·\left|\stackrel{→}{\text{B}}\right|\sin \mathrm{θ}\hat{k}=(\stackrel{→}{\text{A}}×\stackrel{→}{\text{B}})\hat{k}\end{array}$

Now, by the definition of vector product,

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

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

Therefore,

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

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

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

Hence, proved.