When is this vector valued function pointing towards the origin?

Question

"A fighter plane, which can shoot a laser beam straight ahead, travels along the path $\mathbf{r}(t) = \langle 5 - t, 21 - t^2, 3 -\frac{1}{27}t^3\rangle$. Show that there is precisely one time $t$ at which the pilot can hit a target located at the origin." I tried solving $\mathscr l(s) = -\mathbf r(t)$ for t and s, where $\mathscr l(s)$ is the tangent vector of $\mathbf r$ when $\mathbf r$ is pointing towards the origin, but I got two answers for $t%$, one of which was the answer in the back of the book ($t = 3$), and I don't even know what the heck the numbers I got for $s$ were supposed to be... What do I do?

EDIT: I miscopied the question, but now it is correct.

Answer 1

You need $\mathbf{v}(t) = \dot{\mathbf{r}}(t) = \left\langle -1,\, -2t,\, -\frac{1}{9}t^2\right\rangle$ to point to the opposite direction of $\mathbf{r}(t)$ (from the fighter plane to the origin). So just solve $\dot{\mathbf{r}}(t) = k\mathbf{r}(t)$ for $k<0$, which is a system of $3$ equations with $2$ variables ($k$ and $t$).

Answer 2

A hint:

You want that ${\bf r}(t)\times \dot {\bf r}(t)={\bf 0}$ and that ${\bf r}(t)\cdot \dot {\bf r}(t)<0$. The first equation says that $\dot{\bf r}(t)$ is in line with the vector ${\bf r}(t)$ pointing from ${\bf 0}$ to ${\bf r}(t)$, or vice versa, and the second equation says that the shooting direction should point towards the origin and not away from it.

Note that for an "arbitrary" flight travel you cannot expect such a happy moment.

Answer 3

At time $t$, the plane is pointing in the direction determined by the vector $${\bf r}'(t) = \left\langle -1, -2 t, -\tfrac{1}{9} t^2 \right\rangle.$$ This is toward (or away from) the origin if this is parallel to the position ${\bf r}(t)$, that is, if $${\bf r}(t) \times {\bf r}'(t) = 0.$$ This gives a system of three polynomial equations in $t$.

For each solution $t_0$ of this system, it remains to check whether the plane is pointing toward or away from the origin, that is, whether ${\bf r}(t_0)$ and ${\bf r}'(t_0)$ are pointing (respectively) in the opposite or same direction. We can check this by inspection, or algebraically by checking whether $${\bf r}(t_0) \cdot {\bf r}'(t_0)$$ is negative. (In fact, we can just check whether the product of, e.g., the first components of the two vectors is negative.)