Calculating Volume of Parallelpiped

[Lancelot59]

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.

 

[Je m'appelle]

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

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

 

[Lancelot59]

I guess that works. Thanks for the help.

 

[HallsofIvy]

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.