Do these relationships hold true for all vectors? $(u \times v)·w$ is the negated version of $(u \times w)·v$  and vice versa
$u \times (v + w) = (u \times v) + (u \times w)$
I believe that these both hold true for all vectors, even when I substitute a zero vector into either u, v or w. But I am not 100% confident, could somebody please clarify. Thanks
 A: Yes.
The first can be proven from anticommutativity of the cross product and associativity of the triple product:
$$(u\times v)\cdot w = u\cdot(v\times w) = u\cdot(-w\times v) = -u\cdot(w\times v) = -(u\times w)\cdot v$$
The second is a statement of "linearity", that the cross product distributes over addition.
In both cases, if $u = 0$, then all the products are $0$.
A: Let's start with
$u \times (v + w) = (u \times v) + (u \times w)$
This is true, and this is one of the most important properties of vector's cross-product operation: it is linear. I guess the easiest way to make sure it's true is to write cross-product formula in coordinates form. F.e. if we have $a = b \times (c + d)$ than:
$a_x = b_y (c_z+d_z) - b_z (c_y+d_y) = (b_y c_z - b_z c_y) + (b_y d_z - b_z d_y) $
and it's the same as sum of $x$ coordinates of two vectors: $b \times c$ and $b \times d$.
Situation is similar with other coordinates, and we have $a = b \times (c + d) = b \times c + b \times d$
Now back to your first equality.
$(u \times v)·w = - (u \times w)·v$
This is also correct!
Note, that $|u \times v|$ is the area of the parallelogram formed by vectors $u$ and $v$. The vector $u \times v$ itself is orthogonal to this parallelogram. That means that $|(u \times v)·w|$ is a volume of parallelepiped formed by vectors $u$, $v$ and $w$! So it must be $$|(u \times v)·w| = |(u \times w)·v| = V$$
Actually $(u \times v)·w$ is a volume of oriented parallelepiped - it can be positive or negative depending on vectors and changes sign if you change the direction of one of the vector's to the opposite (very convenient property if you think about it!). It also changes sign if you switch any two vectors with each other. I'll leave the proof of this last property to you.
