Does Changing Vector Components Affect Parallelepiped Volume?

[Rotnort]

Homework Statement

The volume of a parallelepiped defined by the vectors w, u, \text{ and }v, \text{ where } w=u \times v is computed using:
V = w \cdot (u \times v)

However, if the parallelepiped is defined by the vectors w-u, u, \text{ and }v, \text{ where } w=u \times v instead, the volume remains the same. Why? Can this be proven mathematically?

The Attempt at a Solution

I can visualise the subtraction of the vector v as not modifying the magnitude of the length of the cross product, rather the horizontal span of the parallelepiped itself. However, I don't know how to prove this without numerical computation.

 

[Ray Vickson]

Homework Statement

The volume of a parallelepiped defined by the vectors w, u, \text{ and }v, \text{ where } w=u \times v is computed using:
V = w \cdot (u \times v)

However, if the parallelepiped is defined by the vectors w-u, u, \text{ and }v, \text{ where } w=u \times v instead, the volume remains the same. Why? Can this be proven mathematically?

The Attempt at a Solution

I can visualise the subtraction of the vector v as not modifying the magnitude of the length of the cross product, rather the horizontal span of the parallelepiped itself. However, I don't know how to prove this without numerical computation.

$$ \mathbf{(w-u) \cdot (u \times v)} = \mathbf{w \cdot (u \times v)} - \mathbf{ u \cdot (u \times v)}$$
What is $\mathbf{u\cdot (u \times v)} ?$

 

[Rotnort]

$$ \mathbf{(w-u) \cdot (u \times v)} = \mathbf{w \cdot (u \times v)} - \mathbf{ u \cdot (u \times v)}$$
What is $\mathbf{u\cdot (u \times v)} ?$

I see, very simple. Aside, is my geometrical justification valid?

 

[Ray Vickson]

I see, very simple. Aside, is my geometrical justification valid?

I don't know: I did not understand what you were saying or what you were trying to do.

 

[Rotnort]

I don't know: I did not understand what you were saying or what you were trying to do.

Ok. I am curious: how would you describe this property geometrically?

 

[fresh_42]

One way to look at it is $\mathbf{w}\cdot(\mathbf{u}\times \mathbf{v})= \det(\mathbf{w},\mathbf{u},\mathbf{v})=\operatorname{Vol}(\mathbf{w},\mathbf{u},\mathbf{v})$.
Now your equation becomes $\det(\mathbf{w},\mathbf{u},\mathbf{v}) = \det(\mathbf{w-u},\mathbf{u},\mathbf{v})$ which is a simple column vector manipulation that doesn't change the linear equation system, esp. the determinant. Some insights on the various vector products can also be found here: https://arxiv.org/pdf/1205.5935.pdf

Unfortunately, the English Wikipedia entry https://en.wikipedia.org/wiki/Triple_product#Geometric_interpretation is a bit short on the geometric interpretation. The Spanish and German versions are better, if you like to switch languages and see if you can understand them. They basically derive the formula of the Volume by
$$
\operatorname{Vol}(\mathbf{u},\mathbf{v},\mathbf{w}) = \operatorname{A}(\mathbf{u},\mathbf{v}) \cdot h_\mathbf{w} = \operatorname{A}(\mathbf{u},\mathbf{v}) \cdot |\mathbf{w}| \cdot \cos(\sphericalangle(\mathbf{u} \times \mathbf{v}), \mathbf{w}) = |\mathbf{u} \times \mathbf{v}| \cdot \mathfrak{n}_{\mathbf{u} \times \mathbf{v}}\cdot \mathbf{w} = |(\mathbf{u} \times \mathbf{v})\cdot \mathbf{w}|
$$
which is the geometric description.

If the vectors $\{\mathbf{u},\mathbf{v},\mathbf{w}=\mathbf{u}\}$ are coplanar, then $\sphericalangle((\mathbf{u} \times \mathbf{v}),\mathbf{u}) = \frac{\pi}{2}$, because $\mathbf{u} \times \mathbf{v}$ is perpendicular to $\mathbf{u}$ and consequently $\operatorname{Vol}(\mathbf{u},\mathbf{v},\mathbf{u}) = 0$.