Hello everyone we have exams tomorrow and i am practising vectors and i wanted some help here. Finding the area of the parallelogram spanned by vectors <-1,0,2> and <-2,-2,2> I have not tried anything since I have no idea. I consider this as revision I have looked at several examples but most are complex and so i want to be helped on this one.
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$\begingroup$This is what I could have done.Just a hint towards exams.
$$Area = {\vert\vec a \times \vec b\vert}$$
$${\vert\vec a \times \vec b\vert}$$= $$ \begin{vmatrix} i & j & k \\ -1 & 0 & 2 \\ -2 & -2 & 2 \\ \end{vmatrix}$$now that we got our tangent vector specified,$$ \begin{vmatrix} 0 & 2 \\ -2 & 2 \\ \end{vmatrix} i -\begin{vmatrix} -1 & 2 \\ -2 & 2 \\ \end{vmatrix} j + \begin{vmatrix} -1 & 0\\ -2 & 2 \\ \end{vmatrix} k$$∴ your equation of line should be;$$ 4i-(-2+4)j+2k = 4i+2j+2k$$
Thus Area = $${\vert\vec a \times \vec b\vert} = \sqrt{4^2 + 2^2 +2^2}=\sqrt{24}$$ Hope this helps you.
$\endgroup$ $\begingroup$The area of the parallelogram spanned by two vectors is the magnitude of the cross-product. The cross product is $<4,-2,2>$, having magnitude $\sqrt{24}$.
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