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Home  >>  TN XII Math  >>  Vector Algebra
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Prove by the vector method, thet the parallelogram on the same base and between the same parallels are equal in area.

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  • By $ \Delta$ law of vectors if $ \overrightarrow a+\overrightarrow b=\overrightarrow c \: or \: \overrightarrow a+\overrightarrow b=-\overrightarrow c$ then the vectors form the sides of a $ \Delta$
  • If $ \overrightarrow a \: and \: \overrightarrow c$ are parallel vectors then $ \overrightarrow a \times \overrightarrow c = \overrightarrow 0$
  • The area of a parallelogram whose sides are $ \overrightarrow a \: and \overrightarrow b \: is \: |\overrightarrow a \times \overrightarrow b|$
Step 1
Let $ABCD \: and \: ABC'D' $ be two parallelograms on the same base AB and between the same parallels $ l \: and \: m$
Step 2
The vector area of parallellogram $ABCD = \overrightarrow {AB} \times \overrightarrow {AD}$
$ = \overrightarrow {AB} \times (\overrightarrow {AD'} \times \overrightarrow {D'D})$
$ = \overrightarrow {AB} \times \overrightarrow {AD'} + \overrightarrow {AB} \times \overrightarrow {D'D}$
$ \overrightarrow {AB} \times \overrightarrow {AD'} + \overrightarrow 0 \: \: ( \because \overrightarrow {AB} \: and \:\overrightarrow {D'D}$ are parallel )
$ = \overrightarrow {AB} \times \overrightarrow {AD'}$
= vector area of parallelogram $ABC'D'$. Hence proved
answered Jun 6, 2013 by thanvigandhi_1
 

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