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Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
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Prove that the area of a paralellogram with diagonals $\bar{a}\;and\;\bar{b}\;is\;\frac{1}{2}|\bar{a} \times \bar{b}|$

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  • Triangular law of addition: In a $\Delta\:PQR,\:\overrightarrow {PQ}+\overrightarrow {QR}=\overrightarrow {PR}$
  • In a parallelogram the diagonals bisect each other.
  • Area of a Parallelogram PQRS=$|\overrightarrow {PQ}\times\overrightarrow {PS}|$
  • $\overrightarrow a\times\overrightarrow a=\overrightarrow 0$
  • $\overrightarrow a\times\overrightarrow b=-\overrightarrow b\times\overrightarrow a$
Let PQRS be a parallelogram with $\overrightarrow {PR}=\overrightarrow a$ and $\overrightarrow {QS}=\overrightarrow b$ as diagonals.
Let the diagonals PR and QS meet at the point T.
Since we know that the diagonals bisect,
$\overrightarrow {PT}=\overrightarrow {TR}=\frac{1}{2}\overrightarrow {PR}=\frac{1}{2}\overrightarrow a\:\:and$
$\overrightarrow {QT}=\overrightarrow {TS}=\frac{1}{2}\overrightarrow {QS}=\frac{1}{2}\overrightarrow b$
In $\Delta \:PQT$, using triangular law of addition we get
$\overrightarrow {PQ}=\overrightarrow {PT}+\overrightarrow {TQ}$
We know that $\overrightarrow {TQ}=-\overrightarrow {QT}$.
$\Rightarrow\:\overrightarrow {PQ}=\overrightarrow {PT}-\overrightarrow {QT}=\frac{1}{2}\overrightarrow a-\frac{1}{2}\overrightarrow b$
Similarly in $\Delta \:PST$, using triangular law of addition we get
$\overrightarrow {PS}=\overrightarrow {PT}+\overrightarrow {TS}=\frac{1}{2}\overrightarrow a+\frac{1}{2}\overrightarrow b$
$|\overrightarrow {PQ}\times\overrightarrow {PS}|=|(\frac{1}{2}\overrightarrow a-\frac{1}{2}\overrightarrow b)\times(\frac{1}{2}\overrightarrow a+\frac{1}{2}\overrightarrow b)|$
$=\frac{1}{4}|(\overrightarrow a-\overrightarrow b)\times(\overrightarrow a+\overrightarrow b)|$
$=\frac{1}{4}\big|\overrightarrow a\times\overrightarrow a+\overrightarrow a\times\overrightarrow b-\overrightarrow b\times\overrightarrow a-\overrightarrow b\times\overrightarrow b\big|$
We know that $\overrightarrow a\times\overrightarrow a=\overrightarrow 0$ and $\overrightarrow a\times\overrightarrow b=-\overrightarrow b\times\overrightarrow a$
$\Rightarrow\:|\overrightarrow {PQ}\times\overrightarrow {PS}|=\frac{1}{4}\big|\overrightarrow 0+\overrightarrow a\times\overrightarrow b+\overrightarrow a\times\overrightarrow b+\overrightarrow 0\big|$
$=\frac{1}{4}\big|2\overrightarrow a\times\overrightarrow b\big|$=$\frac{1}{2}|\overrightarrow a\times\overrightarrow b|$
We know that area of a Parallelogram PQRS=$|\overrightarrow {PQ}\times\overrightarrow {PS}|$
$\Rightarrow\:Area \:=\:\frac{1}{2}|\overrightarrow a\times\overrightarrow b|$.
Hence proved.
answered Apr 20, 2013 by rvidyagovindarajan_1

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