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I Have a parallelogram with the vertices $(1,0,1),(3,1,4),(0,2,9)$ and $(-2,1,6)$

I need to find the area of this parallelogram

My Attempt: I understand My end goal is to find the determinant of a matrix formed by these coordinates. So I first translated the whole parallelogram to the origin by subtracting each vertex by $(1,0,1).$

I am left with $(0,0,0),(2,1,3),(-1,2,8)$ and $ (-3,1,5)$ I can ignore the origin point now, but I believe I am suppose to only use 2 vertices. How do I pick these two vertices, I understand I could draw it out, but is there an easier way? Also can this question be done without translating the vertices?

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  • $\begingroup$ A little bit of magic: if $\mathbf{a}$ and $\mathbf{b}$ are vectors in $\mathbb R^3$, then the area of the parallelogram formed by $0,\mathbf{a},\mathbf{b},$ and $\mathbf{a+b}$ is the magnitude of the vector product, $|\mathbf{a}\times \mathbf{b}|$. $\endgroup$ – TonyK Mar 8 '17 at 19:05
  • $\begingroup$ Choose the two which add up to the other one. $\endgroup$ – Dylan Mar 8 '17 at 19:06
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Ahh Thank You for the hints above. I Was able to solve it:

Vectors (0,0,0) (2,1,3) (-1,2,8) (-3,1,5)

(-3,1,5)-(2,1,3) = (-1,2,8) So these are are two vertices needed to find the determinant.

The determinant is -2i +19j -5k. Take the magnitude of this, which is sqrt(390)

Thanks for the help!

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