Very simple - Volume of Parallelepiped

Question

I'm trying to understand why the volume of a parallelepiped whos sides are $s,u,w$ is $ V = s \cdot(u \times w)$.

Even the units of measurement don't add up. The length of the vectors $s,u,w$ is measured in centimeters, the volume is measured in cubic cm.

$u\times w$ is a vector. It is a vector that is orthogonal to $u$ and $w$, but still a vector, so its length is again measure in cms. So overall $V=s \cdot(u \times w)$ means that $V$ is equal to the product of $2$ vectors, so the unit of measurement for $V$ is squared centimeters, not cubed.

I'm struggling to understand how can $|u\times w|$ be equal to the area of a parallelogram. That is equivalent to saying "The time it takes for me to solve a problem is the distance between New York and London."

Answer 1

The norm of the vector $u\times v$ is defined as the area of the parallelogram (scroll down to Geometric Definition under The Cross Product if you click that link)with sides $u$ and $v$. Also, as both $u$ and $v$ have units of cm, their product will have units of cm$^2$ -- regardless of the fact that $u\times v$ is a vector. Vectors don't have to have units of length -- they can have whatever units we like.

So if $\|u\times v\|$ is the area of a parallelogram, then the area of the parallelopiped will just be this area times the height of the parallelepiped ("bases times height" is the formula we use here). So because $\|s\|\cos(\theta)$ is the height of the parallelepiped (draw a picture to confirm this for yourself), the volume will just be $\|s\|\cos(\theta)\|u\times v\| = \|s\|\|u\times v\|\cos(\theta)$. But that's just the dot product of $s$ and $u\times v$.

Answer 2

I think the apparent paradox in this question derives from the mistaken belief that a vector has associated units of length. This is incorrect. In three-dimensional space, the vector $(1,2,3)$ could be thought of as having units of ordered triples of centimeters (not the same as centimeters cubed). We could also have the coordinates be centimeters, seconds, liters, but it is still an ordered triple.

Now, the length of a vector $|(1,2,3)|=\sqrt{1^2+2^2+3^2}$ is indeed in centimeters, because each of $1,2,3$ is in centimeters, so when squared is in square centimeters, and then the sum has square root applied, which takes us back to centimeters.

If the units were centimeters, seconds, liters, then the length of the vector no longer has any meaningful units anymore.

Answer 3

The cross product has dimensions of area $ |A|\cdot|B| \sin \theta $. Just because it is a vector, it need not be a straight line always. We say an area is directed quantity. An area or even a volume can be a vector.

How do we accept angles, angular velocities, angular accelerations as vectors? Forces as vectors? Current and voltage as vectors? Each has a physical magnitude and direction.

The triple product is scalar with no direction.

The dimensions of volume of a parallelopiped is 2+1 = 3.