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Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
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What is the area of the parallelogram having diagonals $ 3\bar{i}+\bar{j}-2\bar{k}\;and\;\bar{i}-3\bar{j}+4\bar{k}$?

$\begin{array}{1 1} 5 \sqrt 3 \\ 10 \sqrt 3 \\ 8 \\ 4 \end{array} $

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Toolbox:
  • Area of parallelogram =$\large\frac{1}{2}|$$\overrightarrow a\times\overrightarrow b|$ where $\overrightarrow a\:\:and\:\:\overrightarrow b$ are diagonals of the prarallelogram.
  • $\overrightarrow a\times\overrightarrow b=\left|\begin{array}{ccc}\hat i&\hat j&\hat k\\a_1&a_2&a_3\\b_1&b_2&b_3\end{array}\right|$
Given that $\overrightarrow a=3\hat i+\hat j-2\hat k$ and
$\overrightarrow b=\hat i-3\hat j+4\hat k$ are diagonals of the parallelogram.
We know that$\overrightarrow a\times\overrightarrow b=\left|\begin{array}{ccc}\hat i&\hat j&\hat k\\a_1&a_2&a_3\\b_1&b_2&b_3\end{array}\right|$
$=\left|\begin{array}{ccc}\hat i&\hat j&\hat k\\3&1&-2\\1&-3&4\end{array}\right|$ $=\:\hat i(4-6)-\hat j(12-(-2))+\hat k(-9-1)$
$=-2\hat i-14\hat j-10\hat k$
$|\overrightarrow a\times\overrightarrow b|=\sqrt{(-2)^2+(-14)^2+(-10)^2}$
$=\sqrt{4+196+100}=\sqrt{300}=10\sqrt3$
We know that area of the parallelogram =$\large\frac{1}{2}|$$\overrightarrow a\times\overrightarrow b|$ where $\overrightarrow a\:\:and\:\:\overrightarrow b$ are diagonals of the prarallelogram.
Area of the parallelogram $=\large\frac{1}{2}$$10\sqrt3=5\sqrt3 \approx 8.66$
answered Jan 30, 2014 by balaji
edited Jan 30, 2014 by balaji
 

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