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Find the volume of the parallelepiped with vertices $(0,0,0)$, $(3,0,0)$, $(0,5,1)$, $(3,5,1)$, $(2,0,5)$, $(5,0,5)$, $(2,5,6)$, and $(5,5,6)$.

When in $2D$, I usually rely on the Shoelace Formula to solve these types of questions. How do you do this when it's in $3D$?

David G. Stork
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user406996
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1 Answers1

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You can use something called the "three dimensional scalar product". A parallelepiped is uniquely determined by three vectors $u, v, w$, and the volume is given by $|u\cdot(v\times w)|$. In this case, your three vectors are $u = (3, 0, 0), v = (0, 5, 1), w = (2, 0, 5)$.

You'll see this is similar to what you're used to, because for a 2D parallelogram determined by vectors $u, v$, the area is $u\times v$

Edit Just to be sure - are you familiar with the dot product and cross product? Suppose $u = (u_1, u_2, u_3), v = (v_1, v_2, v_3)$. Then $u\cdot v = u_1v_1+u_2v_2+u_3v_3$, and $u\times v = (u_2v_3-u_3v_2, u_3v_1-u_1v_3, u_1v_2-u_2v_1)$

Harambe
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  • I've figured out how you got the three vectors and I know the scalar product you used. Thanks! – user406996 Mar 25 '17 at 01:03
  • No problem, glad it helped. Whenever I've got a list of coordinates defining a parallelepiped, I always draw it to make sure I've got the right three vectors. – Harambe Mar 25 '17 at 01:06