Area of parallelogram having diagonals {3hat{i}},, + ,,hat{j},, - ,,2hat{k} and hat{i},, - ,,3hat{j}

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3-1.Vectors

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The area of the parallelogram having diagonals ${3\hat i}\,\, + \,\,\hat j\,\, - \,\,2\hat k$ and $\hat i\,\, - \,\,3\hat j\,\, + \;\,4\hat k$ is

A

$14$

B

${\rm{5}}\sqrt {\rm{3}}$

C

${\rm{10}}\sqrt {\rm{3}}$

D

${\rm{20}}\sqrt {\rm{3}}$

Solution

Here $a=3 i+j-2 k, b=i-3 j+4 k$

Thus $a \times b =\left|\begin{array}{ccc} i & j & k \\ 3 & 1 & -2 \\ 1 & -3 & 4\end{array}\right|$$=-2 i-14 j-10 k =-2(i+7 j +5 k )$

Hence area of parallelogram $=\frac{1}{2}| a \times b |=5 \sqrt{3}$

Standard 11

Physics

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