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Home  >>  TN XII Math  >>  Vector Algebra
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Find the area of the triangle whose vertices are $(3, -1, 2), (1 ,-1, -3 ), $and $(4, -3, 1) $

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  • The area of a triangle, two of whose sides are $ \overrightarrow a \: and \: \overrightarrow b \: is \: \large\frac{1}{2} |\overrightarrow a \times \overrightarrow b|$
Step 1
$A(3, -1, 2), \: B(1, -1, -3 ), \: C(4, -3, 1)$
Here vector area of triangle = $ \large\frac{1}{2} (\overrightarrow {AB} \times \overrightarrow {AC})$
$ \overrightarrow {AB} = (1-3)\overrightarrow i+(-1+1)\overrightarrow j+(-3-2)\overrightarrow k$
$ -2\overrightarrow i-5\overrightarrow k$
$ \overrightarrow {AC}= (4-3)\overrightarrow i+(-3+1)\overrightarrow j+(1-2)\overrightarrow k$
$ = \overrightarrow i-2\overrightarrow j-\overrightarrow k$
Step 2
$ \overrightarrow {AB} \times \overrightarrow {AC} = \begin{vmatrix} \overrightarrow i & \overrightarrow j & \overrightarrow k \\ -2 & 0 & -5 \\ 1 & -2 & -1 \end{vmatrix} = (0-10)\overrightarrow i - (2+5)\overrightarrow j+(4-0)\overrightarrow k$
$ = -10\overrightarrow i+7\overrightarrow j+4\overrightarrow k$
Vector area of $ \Delta \: ABC = \large\frac{1}{2}(-10\overrightarrow i+7\overrightarrow j+4\overrightarrow k)$
Area = $ \large\frac{1}{2} \sqrt{100+49+16} = \large\frac{\sqrt{165}}{2}$ sq. units
answered Jun 6, 2013 by thanvigandhi_1
 

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