Find the area of parallelogram of the two vectors (6,0,1,3) and (2,1,3,1). The area is magnitude of the cross product of the two vectors. Right now the only way i was taught to do the cross product is getting the determinant of putting the vectors in and i,j,k matrix. I dont know how to do it with 4 points of the vector.
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3The cross-product formula for area is only applicable in 3D space; indeed, there is no such thing as a cross product in 4D space. So you'll need something more basically geometric than that. Try to figure out how you'd compute the area of a parallelogram in 2D space, for instance, without using cross product. – Semiclassical Dec 05 '16 at 16:55
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base x height ? – Chris Dec 05 '16 at 16:56
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2The area would be $|a||b||\sin\theta|$. – anon Dec 05 '16 at 16:57
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The answer sqrt(46x15-18x18) how does this make sense – Chris Dec 05 '16 at 16:59
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1Since $\sin^2\theta=1-\cos^2\theta$ you can do $|a|^2|b|^2-|a\cdot b|^2$ to get the square of the area. – anon Dec 05 '16 at 17:07
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Thank you much appreciated! – Chris Dec 05 '16 at 17:12
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For: $$ \vec v=(6,0,1,3)^T \qquad \vec u=(2,1,3,1)^T $$ we have: $$\vec v \cdot \vec u= 18 \qquad |\vec v|=\sqrt{46} \qquad |\vec u|=\sqrt{15} $$
so, the angle $\theta$ between the two vectors is such that: $$ \cos \theta= \frac{18}{\sqrt{46}\sqrt{15}} \qquad \sin \theta= \sqrt{1-\cos^2 \theta} $$
And the area is: $$ A= |\vec v| \cdot |\vec u| \sin \theta= \sqrt{46}\sqrt{15}\sqrt{1-\frac{18^2}{46 \cdot 15}} =\sqrt{46\cdot 15 - 18^2} $$
Emilio Novati
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@GCab You're right, however answering questions through comments is discouraged in general. – John Dec 05 '16 at 17:44
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The area of your (v,w)-parallelogram is equal to $\sqrt{Gram(v,w)}=\sqrt{\det(AA^T)}$, with the matrix $A=\begin{bmatrix}v & w\end{bmatrix}$. See, for example, Why is the volume of a parallelepiped equal to the square root of $\sqrt{det(AA^T)}$ or How do I compute the area of this parallelogram.
