There are two lines r=a+tu and r'=b+t'v, where t and t' are scalars. Show that if they intersect, then [v,b,u]=[v,a,u].
I've tried finding the intersection between the lines and working from there, but I'm generally just quite confused about how I am to related this to the scalar triple product.Thank you.
On
Since the directions of the lines are the same ($u$) they only intersect, if they are identical, i.e. if there is some $t$ such that
$$(b-a) = tu$$
This is equivalent to $\{b-a,u\}$ being linearly dependent. If your "scalar triple product" is, by chance, defined through
$$[a,b,c] = (a\times b)\cdot c = \det\pmatrix{\vdots&\vdots&\vdots\\a&b&c\\\vdots&\vdots&\vdots}$$
Then this is in fact equivalent to $[v,b-a,u]=0 \ \forall v\in\mathbb R^3$ and thus by linearity of $\det$
$$[v,b,u]-[v,a,u] = 0 \\
[v,b,u] = [v,a,u] \quad\forall v\in\mathbb R^3$$
Note that the latter condidion is a lot more work to check than if $b-a = tu$ for some $t$.
The equivalence is founded on $[a,b,c] = 0 \Leftrightarrow \{a,b,c\}$ is linearly dependent.
I suspect that the lines should be $\mathbf{r}_1(s) = \mathbf{a} + s\mathbf{u}$ and $\mathbf{r}_2(t) = \mathbf{b} + t\mathbf{v}$. I edited the question to reflect this, and my answer below applies to the edited version.
If the lines intersect, then they are coplanar -- they lie in some common plane $P$. Then the three vectors $\mathbf{u}$, $\mathbf{v}$, and $\mathbf{b} - \mathbf{a}$ all lie in the plane $P$. The vector $\mathbf{u} \times \mathbf{v}$ is perpendicular to the plane $P$, so $(\mathbf{u} \times \mathbf{v}) \cdot (\mathbf{b} - \mathbf{a}) = 0$, and the result then follows immediately.