If two lines are skew, then $\overrightarrow{PQ}\cdot\left(\mathbf{u}\times\mathbf{v}\right)\neq0$? Suppose two lines are described by the vector equations:
$$\mathbf{r}=\overrightarrow{OP}+\lambda\mathbf{u} \hspace{10pt} \text{and} \hspace{10pt}\mathbf{r}=\overrightarrow{OQ}+\lambda\mathbf{v} \hspace{10pt} \left(\lambda\in\mathbb{R}\right),$$
where $P$ and $Q$ are points and $\mathbf{u}$ and $\mathbf{v}$ are non-zero vectors (called the lines' direction vectors).
I believe the following claim is true:
Claim. If $\overrightarrow{PQ}\cdot\left(\mathbf{u}\times\mathbf{v}\right)\neq0$, then the two lines are skew.
But is its converse true:
Conjecture. If the two lines are skew, then $\overrightarrow{PQ}\cdot\left(\mathbf{u}\times\mathbf{v}\right)\neq0$.

(Definition. Two lines are skew if they do not intersect and are not parallel.
Definition. Two lines are parallel if their direction vectors can be written as non-zero multiples of each other.)
 A: The quantity $\overrightarrow{PQ} \cdot (\mathbf{u} \times \mathbf{v})$ is known as the scalar triple product. It corresponds with the determinant of the matrix formed by putting the vectors as rows (or columns, equivalently). As such, it is $0$ if and only if the vectors are linearly dependent.
Note that, if two lines are skew, then $\mathbf{u}$ is independent from $\mathbf{v}$, that is, neither is a scalar multiple of each other. Suppose for the sake of contradiction that $\overrightarrow{PQ} \cdot (\mathbf{u} \times \mathbf{v}) = 0$. Then the vectors are linearly dependent, hence
$$\overrightarrow{PQ} = a \mathbf{u} + b \mathbf{v}$$
for some $a, b \in \mathbb{R}$. Consider the plane $\Pi$, going through $P$, generated by directions $u$ and $v$. That is,
$$\Pi = \left\lbrace \overrightarrow{OP} + \lambda \mathbf{u} + \mu \mathbf{v} : \lambda, \mu \in \mathbb{R}\right\rbrace.$$
Then both lines lie in $\Pi$. Obviously the first line lies in $\Pi$ because we can choose $\mu = 0$. The second line lies in $\Pi$ because
$$\overrightarrow{OQ} + \lambda \mathbf{v} = \overrightarrow{OP} + \overrightarrow{PQ} + \lambda \mathbf{v} = \overrightarrow{OP} + a\mathbf{u} + (\lambda + b) \mathbf{v} \in \Pi.$$
The lines are coplanar, hence they follow the Euclidean dichotomy: they intersect or they're parallel. Either way, they're not skew, which is a contradiction.
