If I have an object-1 at $p_1$ with velocity $v_1$ and object-2 at $p_2$ with velocity $v_2$. How would you determine if the objects are moving towards each other?
The requirement is that it must be proved in a single expression.
All I know is that the dot product must be involved.
On
Hint:
I suppose that ''moving towards each other'' means that the velocities are collinear and opposed, so this implies that the their dot product is :$\vec v_1 \cdot \vec v_2=-|\vec v_1||\vec v_2|$
On
I suppose that "moving towards each other" means that the distance between the objects is decreasing.
The velocity of object 2 relative to object 1 is given by $v := v_2 - v_1$.
The displacement of object 2 from object 1 is given by $d:= p_2 - p_1$.
We may simply take the dot product $v \cdot d$. If the result is positive, then the objects are moving away from each other. If the result is negative, then the objects are moving towards each other. If the result is $0$, then the distance is (at that instance) not changing.
The objects will only collide (assuming that the velocities are constant over time) if $v$ is parallel to $d$, which is to say that $v$ is a multiple of $d$. Of course, we must additionally have that the objects are moving towards each other.
On
They will be on a collision course, i.e. heading towards each other if the relative velocity is parallel to the initial relative displacement.
This is because at time $t$ the position of the first particle is $r_1=p_1+tv_1$ and the position of the second particle is $r_2=p_2+tv_2$. They will be in the same place at the same time if there is a $t>0$ such that $$r_1=r_2 \Rightarrow p_1+tv_1=p_2+tv_2 \Rightarrow p_1-p_2=t(v_2-v_1)$$
You could, if you like, write this as a cross-product $$(p_1-p_2)\times(v_2-v_1)=0$$
On
If you mean will they collide, it means that $(\vec v_2 - \vec v_1)t + (\vec p_2 - \vec p_1) = \vec 0$ for some positive $t$.
If you mean are they "aimed at each other", then you must also impose that $\frac{\vec v_1}{|\vec v_1|} - \frac{\vec v_2}{|\vec v_2|} = \vec 0$.
In cases where they're not "aimed at each other", one object could be pointed away from the other object, but the latter could be fast enough to catch up to it. The more general situation within that case is that two objects are pointed in arbitrary directions and happen to have a point of collision.
One idea that simplifies this kind of problem greatly is to take the viewpoint of an observer sitting on one of the object. Suppose we put an observer on Object 1.
In that frame of reference, Object 1 is at the origin of coordinates (the observer's position) and is not moving at all, but Object 2 is (possibly) somewhere else and is (possibly) moving.
Mathematically, we change coordinates so that Object 1 is at the origin by subtracting $p_1$ from all positions. So Object 1 is at $p_1 - p_1 = 0$, and Object 2 is at $p_2 - p_1$. We also change all velocities by subtracting $v_1$, so object 1 has velocity $v_1 - v_1 = 0$, and Object 2 has velocity $v_2 - v_1$.
Using the observer's frame of reference, then, does the observer see Object 2 approaching or receding?