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Home  >>  TN XII Math  >>  Vector Algebra
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Find the area of the parallelogram determined by the sides $\overrightarrow{i}+\overrightarrow{2j}+\overrightarrow{3k}$ and $-\overrightarrow{3i}-\overrightarrow{2j}+\overrightarrow{k}$

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  • The area of a parallelogram whose sides are $ \overrightarrow a \: and \overrightarrow b \: is \: |\overrightarrow a \times \overrightarrow b|$
Vector area of the parallelogram = $ \overrightarrow a \times \overrightarrow b$
where $ \overrightarrow a = \overrightarrow i+2\overrightarrow j+3\overrightarrow k, \: \: \: \overrightarrow b = -3\overrightarrow i-2\overrightarrow j+\overrightarrow k$
$ \overrightarrow a \times \overrightarrow b = \begin{vmatrix} \overrightarrow i & \overrightarrow j & \overrightarrow k \\ 1 & 2 & 3 \\ -3 & -2 & 1 \end{vmatrix} = (2+6)\overrightarrow i-(1+9)\overrightarrow j+(-2+6)\overrightarrow k$
$ = 8\overrightarrow i-10\overrightarrow j+4\overrightarrow k$
Area $ = \sqrt{64+100+16} = \sqrt{180} = 6\sqrt 5 $ sq. units
answered Jun 6, 2013 by thanvigandhi_1
 

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