Question

Prove by vector method that the parallelograms on the same base and between same parallels are equal in area.

Answer 1

Let $\overrightarrow{AB}=\overrightarrow{a}and\overrightarrow{AD}=\overrightarrow{b}$

Area of parallelogram ABCD $=\overrightarrow{a}\times \overrightarrow{b}\to 1$

(Cross product, Since Area= baseXheight)

$=\overrightarrow{a}\left(\overrightarrow{b}\sin \theta \right)$

Now in parallelogram ABB'A

$AB=\overrightarrow{a}$ and $A^{\prime }D=m\overrightarrow{a}$ (let)

($\because$ A'D is parallel to AB)

Consider triangle ADA'

By triangular law of vectors

$\Rightarrow A{A}^{\prime }=m\overrightarrow{a}+\overrightarrow{b}$

$\therefore$ Vector area of ABB'A' $=\overrightarrow{a}\times (m\overrightarrow{a}+\overrightarrow{b})=(\overrightarrow{a}\times m\overrightarrow{a})+(\overrightarrow{a}\times \overrightarrow{b})$

($\because \overrightarrow{a}\times \left(m\overrightarrow{a}\right)=0$, both are parallel, $\therefore \sin 0=0$)

$=0+(\overrightarrow{a}\times \overrightarrow{b})$

$=\overrightarrow{a}\times \overrightarrow{b}\to 2$

$\because \left(1\right)=\left(2\right)$,

Hence proved