Let us fix vectors $r_0$ and $u$ in $\mathbb{R}^3$, where $u\ne0$.
Using the identity $u×(v×w) = (u·w)v−(u·v)w$, for vectors in $\mathbb{R}^3$, show that if $r×u = r_0 ×u$, then there exists $t ∈ \mathbb{R}$ (which you should determine) such that $r = r_0 +tu$.
Hint: take the vector product of both sides of the equation with $u$.
My progress so far..From the hint, I have taken the vector product of both sides with $u \times u$ = $0$ on the right hand side of the equation, leaving me with $0$ = ($u·w)v× u−(u·v)w×u$. I am unsure of what the next step to take is. The next step may be made easier if I have the equation in the form $0 = u×((u.w)v-(u.v)w)$.
Hint 1: Follow the hint.
Hint 2 (Alternative mathod): $(r-r_0)\times u=0$.
See also a Gowers's post about "fake difficulty".