1. ## Parallelepiped Vertices

If vector OA, vector OB, and vector OC are three edges of a parallelpiped where O is (0,0,0), A is (2,4,-2), B is (3,6,1) and C is (4,0,-1) find the coordinates of the other vertices of the parallelpiped.

for this q, i am having trouble drawing the diagram of the parallelpiped

Currently though, I have the following possibilities.. (there are only 3 right?)

D(5,10,0)<<< DB//CA
D(3,-2,0)<<<CB//DA

2. Hello skeske1234
Originally Posted by skeske1234
If vector OA, vector OB, and vector OC are three edges of a parallelpiped where O is (0,0,0), A is (2,4,-2), B is (3,6,1) and C is (4,0,-1) find the coordinates of the other vertices of the parallelpiped.

for this q, i am having trouble drawing the diagram of the parallelpiped

Currently though, I have the following possibilities.. (there are only 3 right?)

D(5,10,0)<<< DB//CA
D(3,-2,0)<<<CB//DA
If you're having trouble drawing it, then don't try! Just think of it as being a boring old cuboid (rectangular box) that's been squashed a bit.

Instead of rectangles, then, the faces are all parallelograms. So if we consider the face containing $O$, $A$ and $B$, the fourth vertex of this parallelogram - $P$, say - is such that:

$\vec{OP} = \vec{OA} + \vec{OB} = \begin{pmatrix} 2\\4\\-2\end{pmatrix}+ \begin{pmatrix} 3\\6\\1\end{pmatrix}=\begin{pmatrix} 5\\10\\-1\end{pmatrix}$

Then deal with the 2 faces containing $O$, $A$ and $C$; and $O$, $B$ and $C$, respectively, in the same way to get, say, vertices $Q$ and $R$.

Finally, we shall need the eighth vertex, $S$, say, whose position vector is the sum of all three vectors $\vec{OA} + \vec{OB}+\vec{OC}$. This vertex will be diagonally opposite $O$.

Can you complete it now?

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# how to find edges of parallelopiped

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