You need three scalar quantities to specify a vector in a three-dimensional space. If you know only the magnitude of the vector and the angle it forms with one axis, those two scalar quantities are not enough to fully determine the vector -- it could be any vector with the given magnitude on a cone around that axis. You need one more quantity to fix the vector; then you can calculate the angles and components you're looking for.
P.S.: Note that the title and body of the question are inconsistent. Specifying a direction in three-dimensional space requires two scalar quantities; since you only have one angle, you don't know the direction of the vector.
The notation in the problem statement seems misleading to me.
We are given symbols $P_1,$ $P_2,$ $V_1,$ $V_2,$ $v_1,$ and $v_2,$
seemingly describing objects $1$ and $2$ in a completely symmetric fashion,
whereas the problem statement actually is not symmetric.
The problem statement implies that $P_1,$ $P_2,$ $V_1,$ $v_1,$ and $v_2$
are all known, with $v_1 = \lVert V_1\rVert.$
The unknown entity is $V_2$; we have the constraint that
$\lVert V_2\rVert = v_2,$ but as long as that constraint is satisfied,
we can choose the direction of $V_2$ freely.
The mathematics of finding a suitable direction for $V_2$ were thoroughly hashed out in the answers to Intersection of two moving objects.
There is an algebraic treatment
which concludes with an equation that the angle $\phi$ (the direction of vector $V_2$) must meet in order to make Object $2$ meet Object $1.$
There are at most two solutions of the equation, but in some cases a solution of the equation gives a value of $\phi$ that causes Object $2$ to move away from Object $1$ (so they will never meet), while in other cases there are two values of $\phi$ that would cause the objects to meet, although only one of those values can be the one that makes them meet in the minimum possible time.
I find a geometric approach gives me more insight on the problem;
in particular, once the geometry of the problem has been set up
in this way
(see this answer
or this answer)
it immediately gives us the equation
$$
v_2 \cos \phi = -v_1 \cos\theta,
$$
where $\phi$ is the angle between $V_2$ and the line $P_2P_1$
and $\theta$ is the angle between $V_1$ and the line $P_2P_1,$
measuring both angles counterclockwise.
(The algebraic treatment comes to an equivalent conclusion.)
Another way to say this is:
In order for the two objects to meet, the component of $V_2$ perpendicular to the line between the objects must exactly cancel the component of $V_1$ perpendicular to the line between the objects.
This condition is necessary but not sufficient to solve the problem.
A condition that is both necessary and sufficient is:
The vector $V_2 - V_1$ must point in the direction from $P_2$ to $P_1.$
Geometrically, we can construct representations of all possible vectors of the form $V_2 - V_1$ (given that $V_1$ is fixed and the magnitude but not the direction of $V_2$ is fixed)
by drawing the vector $-V_1$ and then drawing a circle of radius $v_2$
around the tip of $V_1.$
We can then represent $V_2 - V_1$ by placing its tail at the tail of
$-V_1$ and its head at any point on the circle; the direction from the center of the circle to the chosen point is the direction of $V_2.$
The diagram in this answer
shows the case where $v_1 > v_2$ but where it is still possible to make the objects meet.
In this case the circle intersects the line $P_1P_2$ in two places,
giving two possible choices of a vector $V_2$ that will cause the objects to meet.
The choice that causes the objects to meet in the minimum possible time is the choice for which the magnitude of $V_2 - V_1$ is larger.
(In the diagram in the linked answer, that vector is $CG.$)
When $v_1 > v_2,$ it can also happen that there is no solution to the problem; object $2$ can only "catch" object $1$ if object $1$ is initially reducing its distance to object $2$ quickly enough.
When $v_2 \geq v_1,$ there is always a solution but only one solution
for $V_2$ that allows object $2$ to "catch" object $1,$
so of course that is also the minimum-time solution.
The solution is shown graphically in this answer.
Best Answer
When they collide, the point of impact between the two is along a plane (you can imagine a straight line between the two at impact) The resulting direction of the target ball is perpendicular to that plane. AKA The direction of the target ball is along the line of the center of the cue ball, to the center of the target ball.
(This isnt from my math, this is from my experience in playing pool)