# Calculating Volume of Parallelpiped

I'm doing problems which have me calculate the volumes of parallelpipeds I'm slightly confused with this. I know the formula is:

$$V=\vec{a} \cdot (\vec{b}\times\vec{c})$$

where a and b form the base, with c being the "vertical" side. My issue is that when given three vectors which define this shape, how do I know which order to put the vectors in? For one example I was given the vectors:

<6,3,-1>
<0,1,2>
<4,-2,5>

and after drawing them out I had no clue which was which, so I just decided to use them in order as a, b, and c. It worked.

Then in the next problem I was given four points:
P(2,0,-1)
Q(4,1,0)
R(3,-1,1)
S(2,-2,2)

And told that three sides were defined by PQ, PR, and PS. So after getting those vectors I again just took them in order and got -3. The correct answer is three.

How do I determine which vectors I put in which place in the equation? In the examples they just pick three in order and use them that way.

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## Answers and Replies

All your procedures are correct, you just forgot a little detail, the volume of the Parallelepiped is given by

$$V=|\vec{a} \cdot (\vec{b}\times\vec{c})|$$

That is, the absolute value.

The last answer is |-3| = 3.

By the way, concerning your curiosity regarding the order of the vectors, it does not matter when you're computing the volume, as it is a matter of referential.

See it like this, suppose we have a, b, c for the length, width and height (not in order) of a Parallelepiped, when computing the volume V = abc, it does not matter which is which.

Or mathematically, in vector form

$$V=|\vec{a} \cdot (\vec{b}\times\vec{c})| = |\vec{b} \cdot (\vec{a}\times\vec{c})| = |\vec{c} \cdot (\vec{a}\times\vec{b})|$$

I guess that works. Thanks for the help.

HallsofIvy
Science Advisor
Homework Helper
Yes, changing the order of the vectors just changes the sign which will not affect the absolute value.

It is worth remembering that with $\vec{u}= a\vec{i}+ b\vec{j}+ c\vec{k}$, $\vec{v}= d\vec{i}+ e\vec{j}+ f\vec{k}$, and $\vec{w}= x\vec{i}+ y\vec{j}+ z\vec{k}$

$$\vec{u}\cdot\left(\vec{v}\times\vec{w}\right)=$$$$\left(a\vec{i}+ b\vec{j}+ c\vec{k}\right)\left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ d & e & f \\ x & y & z\end{array}\right|$$$$= \left|\begin{array}{ccc}a & b & c \\ d & e & f \\ x & y & z\end{array}\right|$$

and swapping rows in a determinant only changes the sign.