Rendezvou course for moving boats There is a boat travelling along a straight line on $ \mathbb{R}^2 $ with its location at time t given by $ (t \cos(\theta_a), t \sin(\theta_a) ) $. I'm in a boat that starts at $(x_b, y_b)$ and travels a fixed speed of $v$.  My location at time $t$ is given by $ (vt \cos(\theta_b) + x_b, vt \sin(\theta_b) + y_b ) $.
I can pick my angle $ \theta_b$.  I wish to pick an angle so that I meet up with the other vessel. In other words given $x_b$, $y_b$, $v$ and $ \theta_a$ find $\theta_b$ such that a some time $t_r$ the following equality holds $(t_r \cos(\theta_a), t_r \sin(\theta_a)) = (t_rv \cos(\theta_b) + x_b, t_rv \sin(\theta_b) + y_b) $.
Now clearly there are some combinations of speed and location where there will be no solution.  However in the cases where there is a solution how do I solve it?
 A: Let's call the unit-direction vectors of the lines $\vec a=(\cos\theta_a,\sin\theta_a)$ and $\vec b=(\cos\theta_b,\sin\theta_b)$, resp., and let $\vec p=(x_b,y_b)$ and $\phi$ the angle between $\vec a$ and $\vec p$. We're trying to solve $\vec at=vt\vec b+\vec p$, assume $v>0$.  From here $\vec a-v\vec b$ is parallel to $\vec p$ and must have length $v$.  This gives an easy way to construct $v\vec b$.  
Draw $\vec a$, $\vec p$ and a circle  with center $(0,0)$ and radius $v$.  The parallel to $\vec p$ through $\vec a$ meets that circle in two points, provided that  $v>|\sin(\phi)|$.  These are the candidates for $v\vec b$.
A straightforward computation gives 
$$v\vec b=\vec a-\frac{\vec p}{\|\vec p\|}\left(\cos(\phi)\pm\sqrt{v^2-\sin^2(\phi)}\right).$$
From here it's easy to calculate the time and place where both vessels meet.
A: An alternative approach that may also be useful to understand why there can be two, one, or no solutions - depending on the combination of speed and location - is the following.  Let us call $A $ the point $(t \cos \theta_a, t \sin \theta_a) $ where the two boats meet, and $B $ the point $(x_b,y_b ) $ from which the second boat starts. Also note that, because the first boat reaches $A $ in the time $t $ moving along a line with angle $\theta_a $, it starts at the origin and has velocity equal to $1$.
Since the second boat must run the distance between $B $ and $A $ with velocity $v $ in the time $t $, we can write
$$\sqrt {(t \cos \theta_a-x_b)^2 + (t \sin \theta_a-y_b ) ^2}=vt$$
We can then solve the resulting second degree equation  for $t $, which is the only unknown quantity. This leads to
$$t^2 \cos^2 \theta_a  -  2x_b t \cos \theta_a  +x_b^2 + t^2 \sin^2 \theta_a  -  2y_b t \sin \theta_a  +y_b^2=v^2t^2$$
$$(v^2-1)t^2   + 2 t (x_b \cos \theta_a     +     y_b \sin \theta_a) -x_b^2  -y_b^2=0$$
$$t= \frac {1}{v^2-1} \left(-x_b \cos \theta_a - y_b \sin \theta_a \pm \sqrt {    ( x_b \cos \theta_a     +      y_b \sin \theta_a ) ^2  +(v^2-1)(x_b^2  +y_b^2)} \right)$$
This equation expresses $t $ as a function of the coordinates $(x_b,x_b) $, the velocity $v $ and the angle $ \theta_a $. If there are positive real solutions for $t $, we can get corresponding solutions for the angle $ \theta_b $ by considering that
$$ \theta_b=\arctan_2 \left( t \sin \theta_a-y_b, t \cos \theta_a -x_b\right) $$
where $\arctan_2(q, p)$ is the common variation of the $\arctan $ function, used to identify  the appropriate quadrant of the computed angle and defined as the angle  between the positive $x$-axis and the line connecting the point with coordinates $(p, q)$ with the origin. 
Now let us study the number of valid solutions. First let us set, by simplicity,   $ x_b \cos \theta_a     +      y_b \sin \theta_a=J$. The equation giving $t $ can then be rewritten as
$$t= \frac {1}{v^2-1} \left(-J \pm \sqrt { J^2  +(v^2-1)(x_b^2  +y_b^2)} \right)$$
Since by definition $t $ must be positive, the number of valid solutions obtained for $t $ (and for $ \theta_b $) depends on whether the determinant
$$J^2  +(v^2-1)(x_b^2  +y_b^2)$$
is positive, zero, or negative, and on whether the  resulting  solutions  for $t $ are positive or negative. In particular,  the determinant is $\geq 0$ if
$$v^2 \geq 1- \frac {J^2}{x_b^2  +y_b^2}$$
or equivalently if
$$v^2 \geq \frac {(x_b \sin \theta_a   +    y_b \cos \theta_a)^2}{x_b^2  +y_b^2}$$
Considering that $v $ is by definition positive, we have
$$v \geq \frac {|x_b \sin \theta_a     +      y_b \cos \theta_a|}{\sqrt {x_b^2  +y_b^2}}$$
where it can be easily shown that the quantity in the RHS (let us call it $K $), for  $ x_b^2  +y_b^2 \neq 0$, is $\leq 1$. The value of $K $ can be interpreted as the minimal velocity that is necessary (albeit not sufficient as a condition) for a valid solution to exist.
As a  result, looking again at the formula above that provides the solutions for $t $, we have the following cases:


*

*CASE 1: $v>1$ (second boat faster than the first one). The determinant is positive and the absolute value of the radical is higher than $|J|$,  so that the  two values of $t $ are real, one positive and one negative. Thus, there is only one valid solution. This is also intuitive: whatever the angle $\theta_a $ and the initial position of the second boat, there is no way, for the first boat, to "escape" for an infinite time if the second boat is faster.

*CASE 2: $v=1$ (equal velocity between the first and second boat). In this case, the initial  second degree equation loses its $t^2$ term and reduces to a first degree equation whose solution is
$$ t =\frac {x_b^2  +y_b^2}{ 2 (  x_b \cos \theta_a     +      y_b \sin \theta_a) }= \frac {x_b^2  +y_b^2}{2J }   $$
which is valid only if $J >0$. A very simple example falling in this case is when $\theta_a=0$ (i.e., the first boat moves horizontally along the $x $-axis) and the initial position of the second boat is on the $y=x $ line, e.g. $x_b=y_b=z $. Here, if the boats have equal velocity, the trivial solution is given by a direct vertical movement of the second boat towards the point $(z,0) $, so that both boats perpendicularly converge to this point  with velocity $1$, covering a distance $z $ in the time $z $. Accordingly, the formula above reduces to $ t =(2z^2)/( 2 z )= z$.


*

*CASE 3: $k <v <1$ (second boat slower than the first one, but with velocity higher than the "minimal" velocity $k $). The determinant is still positive and the absolute value of the radical is lower than $| J |$,  so that the  two values of $t $ have concordant sign. Thus, there are two valid solutions if $J$ is positive and no valid solutions if $J$ is negative. A simple scenario for the case in which two valid solutions are obtained is the  trivial case where $ \theta_a=0$ and the first boat runs horizontally rightward along the positive $x $-axis (so that its arrival point $A $ has coordinates $(t,0) $ ), and where $x_b>0$ and  $y_b=0$ (i.e. the starting point of the second boat lies on the positive  $x $-axis as well). In this case, the equation providing the solutions for $t $ has a positive determinant and reduces to
$$t= \frac {1}{v^2-1} (-x_b \pm x_b v)$$


giving the two solutions
$$t= \frac {x_b}{v+1}$$
$$t= -\frac {x_b}{v-1} $$
The first solution, which can also be written as $vt=x_b-t $,  refers to that in which $t <x_b $, i.e.  the second boat runs towards the left along the $x $-axis towards the first boat with velocity $v $, covering the distance  between its starting point and the point $(t,0) $ in the time $t $.  The second solution, which can be written as $vt=t-x_b $, refers to the case in which $t>x_b$,  so that the second boat runs towards the right along the $x $-axis just as the first boat, but with a lower velocity (so that it is reached by the first boat at the point $(0,t) $ in the time $t $). Also note that this second case is clearly valid only for $v <1$. Specular solutions exists in the case that the two boats move along the negative $x $-axis.


*

*CASE 4: $v=k $ (second boat slower than the first one, with velocity equal to the minimal velocity $k $). The determinant is zero, so that there is one valid solution  if $J$ is positive and no valid solution if $J$ is negative.

*CASE 5: $0 <v <k$. The determinant is negative and there are no real solutions for $t $. 
In summary:


*

*there is always one valid solution for $t $ and $\theta_b$ when $v>1$, that is to say if the velocity of the second boat is higher than that of the first boat;

*there is no valid solution when $v<K$, that is to say if the velocity of the second boat is lower than a value representing the "minimal" velocity;

*for intermediate values of $v $ ranging from $K $ to $1$, there can be two, one or no valid solutions, and this can be established by looking at $J $, a quantity easily calculated from $\theta_a $, $x_b $, and $y_b $.
A: \begin{align*}
  |\mathbf{v}_{a}-\mathbf{v}_{b}| &= v \\
  \mathbf{v}_{a}^2-2\mathbf{v}_a \cdot \mathbf{v}_b+\mathbf{v}_b^2 &= v^2 \\
  1-2\cos (\theta_a-\theta_b)v_b+v_b^2 &= v^2 \\
  \cos (\theta_a-\theta_b) &= \frac{1+v_b^2-v^2}{2v_b}
\end{align*}
By triangular inequality,
$$|\mathbf{v}_{a}-\mathbf{v}_{b}|=v \implies
|v_a-v| \le v_{b} \le v_a+v$$
$$|1-v| \le v_{b} \le 1+v$$
For real $v_b$,
\begin{align*}
  4\cos^2 (\theta_a-\theta_b)-4(1-v^2) & \ge 0 \\
  4[v^2-\sin^2 (\theta_a-\theta_b)] & \ge 0 \\
  \sin^2 (\theta_a-\theta_b) & \le v^2
\end{align*}
