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Home  >>  TN XII Math  >>  Vector Algebra
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Find the area of parallelogram $A\;B\;C\;D$ whose vertices are $A(-5 ,2 ,5),B(-3 ,6 ,7),C(4 ,-1 ,5)$ and $D(2 ,-5 ,3)$

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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 $( i.e., \large\frac{1}{2} | \overrightarrow d_1 \times \overrightarrow d_2|)$
Step 1
Area of quadrilateral ABCD = $ \large\frac{1}{2} |\overrightarrow {AC} \times \overrightarrow {BD}|$
$ \overrightarrow {AC} = \overrightarrow {OC}-\overrightarrow {OA}= 9\overrightarrow i-3\overrightarrow j$
$ \overrightarrow {BD}=\overrightarrow {OD} - \overrightarrow {OB} = 5\overrightarrow i-11\overrightarrow j-4\overrightarrow k$
Step 2
Vector area = $ \large\frac{1}{2}(\overrightarrow {AC} \times \overrightarrow {BD}) = \large\frac{1}{2} \begin{vmatrix} \overrightarrow i & \overrightarrow j & \overrightarrow k \\ 9 & -3 & 0 \\ 5 & -11 & -4 \end{vmatrix}$
$ = \large\frac{1}{2} [ 12\overrightarrow i - 36\overrightarrow j+(-99+15)\overrightarrow k]$
$ = \large\frac{1}{2} [ 12\overrightarrow i-36\overrightarrow j-84\overrightarrow k]$
$ = 6 [ \overrightarrow i-3\overrightarrow j-7\overrightarrow k]$
Area $ = 6\sqrt{1+9+49} = 6\sqrt{59}$ sq. units
answered Jun 5, 2013 by thanvigandhi_1
 

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