Question

Find area of the parallelogram whose sides are represented (2î + 4 - 6k) and (i + 2k) :-
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Answer 1

If $\vec{\sf{D_1}}$ and $\vec{\sf{D_2}}$ are two diagonal vectors of a parallelogram, then the area of the parallelogram will be,

$\longrightarrow \vec{\sf{A}}=\pm\mathsf{\dfrac{1}{2}}\left(\vec{\sf{D_1}}\times\vec{\sf{D_2}}\right)$

According to the question,

• $\vec{\sf{D_1}}=\sf{2\,\hat i+4\,\hat j-6\,\hat k}$

• $\vec{\sf{D_2}}=\sf{\hat i+2\,\hat k}$

Then, area is,

$\longrightarrow \vec{\sf{A}}=\pm\mathsf{\dfrac{1}{2}}\left(\sf{\left(2\,\hat i+4\,\hat j-6\,\hat k\right)\times\left(\hat i+2\,\hat k\right)}\right)$

$\longrightarrow \vec{\sf{A}}=\pm\mathsf{\dfrac{1}{2}}\left|\begin{array}{ccc}\sf{\hat i}&\sf{\hat j}&\sf{\hat k}\\\sf{2}&\sf{4}&\sf{-6}\\\sf{1}&\sf{0}&\sf{-2}\end{array}\right|$

$\longrightarrow \vec{\sf{A}}=\pm\mathsf{\dfrac{1}{2}}\left(\sf{(-8+0)\,\hat i-(-4+6)\,\hat j+(0-4)\,\hat k}\right)$

$\longrightarrow \vec{\sf{A}}=\pm\mathsf{\dfrac{1}{2}}\left(\sf{-8\,\hat i-2\,\hat j-4\,\hat k}\right)$

$\longrightarrow\underline{\underline{\vec{\sf{A}}=\pm\mathsf{\left(\sf{4\,\hat i+\,\hat j+2\,\hat k}\right)}}}$

And so its magnitude is,

$\sf{\longrightarrow A=\sqrt{4^2+1^2+2^2}}$

$\sf{\longrightarrow\underline{\underline{A=\sqrt{21}}}}$