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Home  >>  TN XII Math  >>  Vector Algebra
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Find the area of the parallelogram whose diagonals are represented by $\overrightarrow{2i}+\overrightarrow{3j}+\overrightarrow{6k}$ and $\overrightarrow{3i}-\overrightarrow{6j}+\overrightarrow{2k}$

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  • The area of a quadilateral ABCD is $ \large\frac{1}{2} | \overrightarrow {AC} \times \overrightarrow {BD}|$ where AC and BD are the diagonals $( i.e., \large\frac{1}{2} | \overrightarrow d_1 \times \overrightarrow d_2|)$
The diagonals are $ \overrightarrow d_1 = 2\overrightarrow i+3\overrightarrow j+6\overrightarrow k$
$ \overrightarrow d_2 = 3\overrightarrow i-6\overrightarrow j+2\overrightarrow k$
Vector area of the parallelogram = $ \large\frac{1}{2} (\overrightarrow d_1 \times \overrightarrow d_2)$
$ = \large\frac{1}{2} \begin{vmatrix} \overrightarrow i & \overrightarrow j & \overrightarrow k \\ 2 & 3 & 6 \\ 3 & -6 & 2 \end{vmatrix}$
$ = \large\frac{1}{2} [ (6+36)\overrightarrow i - (4-18)\overrightarrow j+ (-12-9)\overrightarrow k]$
$ = \large\frac{1}{2} [ 42\overrightarrow i + 14\overrightarrow j -21\overrightarrow k ]$
$ = \large\frac{7}{2} [ 6\overrightarrow i+2\overrightarrow j-3\overrightarrow k]$
Area $= \large\frac{7}{2} \sqrt{36+4+9} = \large\frac{49}{2} $ sq. units
answered Jun 5, 2013 by thanvigandhi_1
 

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