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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Find the area of the parallelogram whose diagnols are represented by the vectors $ \overrightarrow d_1 = 3\hat i +\hat j -2\hat k\: and \: \overrightarrow d_2 = \hat i -3\hat j +4\hat k.$

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  • Area of the parallelogram is $\large\frac{1}{2}$$\mid d_1\times d_2\mid$
Step 1:
Given :
$\overrightarrow d_1=3\hat i+\hat j-2\hat k$
$\overrightarrow d_2=\hat i-3\hat j+4\hat k$
$\overrightarrow {d_1}\times\overrightarrow {d_2}=\begin{vmatrix}\hat i&\hat j&\hat k\\3 &1&-2\\1 &-3&4\end{vmatrix}$
$\qquad\;\;\quad=\hat i(4-6)-\hat j(12+2)+\hat k(-9-1)$
$\qquad\;\;\quad=-2\hat i-14\hat j-10\hat k$
Step 2:
$\mid\overrightarrow {d_1}\times \overrightarrow {d_2}\mid=\sqrt{(-2)^2+(-14)^2+(-10)^2}$
$\qquad\qquad=\sqrt{300}$
Area of the parallelogram is $\large\frac{1}{2}$$\mid d_1\times d_2\mid$
$\Rightarrow \large\frac{1}{2}$$\sqrt{300}$ sq.units
$\Rightarrow \large\frac{10\sqrt 3}{2}$
$\Rightarrow 5\sqrt 3$ sq.units
answered Nov 7, 2013 by sreemathi.v
 

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