I was asked to find the area of the parallelogram formed from these two sides: <-2,1> and <1,3>.
How can I find the cross-product if it is not in $\mathbb R^3$?
This picture is of my work:
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J. W. Tanner
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Burt
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Hint: the area will be the same as the area of the parallelogram formed by the vectors $\langle -2, 1, 0 \rangle$ and $\langle 1, 3, 0 \rangle$. – user744868 Mar 13 '20 at 01:35
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@user744868 So in order to solve I should just add in the zeros – Burt Mar 13 '20 at 01:36
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yes, and $-6-1=-7$, not $<-6,-1>$ – J. W. Tanner Mar 13 '20 at 01:36
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@J.W.Tanner To do this kind of problem can I just find the determinant of the 2x2 matrix and say that is the magnitude? That works this time. – Burt Mar 13 '20 at 01:39
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2yes, the determinant of a $2\times2$ matrix gives the magnitude of the parallelogram area – J. W. Tanner Mar 13 '20 at 01:45
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It’s right there in the image that you pasted: the area is equal to the determinant of $[\mathbf a\ \mathbf b]$. – amd Mar 13 '20 at 04:39
2 Answers
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You can embed the vectors $\vec a = \langle -2,1\rangle$ and $\vec b= \langle 1,3\rangle$ into $\mathbb{R}^3$ via $ \vec a = \langle -2,1,0\rangle$ and $\vec b = \langle 1,3,0\rangle$. Here, I am using the same notations $\vec a$ and $\vec b$ to represent the embedded vectors. Then $$ |\vec a\times \vec b| = \left| \begin{array}{ccc} i & j & k \\ -2& 1 & 0 \\ 1 & 3 & 0 \\ \end{array} \right| = |-6k-k| = |-7k| = |-7||k| = 7 $$ since $i$, $j$, and $k$ are the standard unit vectors.
Mee Seong Im
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First of all, you could embed the vectors in $\mathbb R^3$ as $(-2,1,0)$ and $(1,3,0)$.
Second of all, the magnitude of the cross-product is $|(0,0,-6-1)|=7.$
J. W. Tanner
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I see - so my mistake was that the discriminant of the 2x2 matrix is the magnitude. – Burt Mar 13 '20 at 01:37
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