A particle was at point $P_1$ at time $t_1$ and is moving at the constant velocity $\vec{v}_1$. Another particle was at $P_2$ at $t_2$ and is moving at the constant velocity $\vec{v}_2$. How close did the particles get to each other and at what time? What conditions are needed for a collision?
[Math] Two particles moving at consant velocity: how close do they get
physicsvectors
Related Solutions
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.
Mathematical Model
It is convenient to use vectors here. For particle $A$ we have: $$ r_A(t) = r_A(0) + v_A t = (x_A(0), 25) + (3, 0) t = (x_A(0) + 3t, 25) $$ where the text gives $x_A(0) = 0$.
For particle $B$ we have
$$
a_B(t) = a \\
v_B(t) = v_B(0) + a t = at \\
r_B(t) = r_B(0) + \frac{1}{2}a t^2
= \frac{1}{2}a t^2
= ((1/2)\lVert a \rVert \sin(\theta) t^2,(1/2)\lVert a \rVert \cos(\theta) t^2)
$$
The above image shows the trajectory of $A$ (green) and of $B$ (blue) and where the trajectories intersect at point $X$. Also the angle is given (named $\alpha$ instead of $\theta$).
You can fiddle with a live version here.
Calculating the Collision Point
For a collision we need $r_A(t) = r_B(t) = X$ for some $t$.
The intersection point $X$ is $$ X = (\xi, 25) $$ with $\tan(\theta) = \xi / 25 \iff \xi = 25 \tan(\theta)$.
$A$ arrives at $X$ if $$ t = (25/3) \tan(\theta) \quad (1) $$ $B$ arrives at $X$ if $$ \frac{1}{2} \lVert a \rVert \sin(\theta) t^2 = 25 \tan(\theta) \quad (2) \\ \frac{1}{2} \lVert a \rVert \cos(\theta) t^2 = 25 \quad (3) $$ Alas equation $(2)$ and $(3)$ are equivalent, division of both sides of $(2)$ by $\tan(\theta)$ gives $(3)$.
Inserting $(1)$ into $(3)$ gives $$ \frac{1}{2} \lVert a \rVert \cos(\theta) \left(\frac{25}{3}\right)^2 \tan^2(\theta) = 25 \iff \\ 25 \lVert a \rVert \sin^2(\theta) = 18 \cos(\theta) \quad (4) $$ To limit the solutions we should note that $\theta \in (0, \pi/2)$.
Algebraic Solution
Substituting (see Tangent half-angle formulas) $$ t = \tan(\theta/2) \\ \sin(t) = \frac{2t}{1+t^2} \\ \cos(t) = \frac{1-t^2}{1+t^2} \\ $$ (note: this $t$ is not the time $t$ from further above, but just used for this intermediate calculation of $\theta$ from equation $(4)$) we get $$ 25 \lVert a \rVert \left(\frac{2t}{1+t^2} \right)^2 = 18 \frac{1-t^2}{1+t^2} \iff \\ 25 \lVert a \rVert 4t^2 = 18(1-t^2)(1+t^2) = 18(1-t^4) \\ t^4 + \frac{50}{9} \lVert a \rVert t^2 - 1 = 0 \quad (5) $$ Introducing $s = t^2$ we get $$ s^2 + \frac{50}{9} \lVert a \rVert s - 1 = 0 \quad (6) $$ which has the solutions $$ s = \frac{\pm \sqrt{81+25^2 \lVert a \rVert^2}-25\lVert a \rVert}{9} $$ we discard the negative solution and apply $25 \lVert a \rVert = 11$ and get $$ s = \frac{\sqrt{202} - 11}{9} \\ t = \frac{\sqrt{\sqrt{202}-11}}{3} \\ \theta = 2 \arctan{\frac{\sqrt{\sqrt{202}-11}}{3}} \approx 1.0771 \approx 61.714^\circ $$ Inserting in equation $(1)$ gives $$ t \approx (25/3) \tan(1.0771) \approx 15.485 \, \text{s} $$
Best Answer
There's collision at time t if : $\vec{OP_1}+(t-t_1)\cdot \vec{V_1}=\vec{OP_2}+(t-t_2)\cdot \vec{V_2}$, where O is the origin.
If not, $\vec{OP_1}+(t-t_1)\cdot \vec{V_1}-(\vec{OP_2}+(t-t_2)\cdot \vec{V_2})=\alpha\cdot\vec{V_1}\wedge\vec{V_2}+\beta(\vec{V_1}+\vec{V_2}) $
Therefore you have a system with 3 equations (projection on $\vec{x},\vec{y},\vec{z}$) and 3 unknowns $(t,\alpha, \beta)$
EDIT
By the way searching for numerical solutions, I found this topic which is exactly the
same problem : here