当一个向量的方向未知时，两个向量之间的交点

Question

Problem: I have two vectors. I know the starting point of one vector, its direction, its magnitude. I know the starting point of the other vector and its magnitude. I need to find the direction of second vector as well as the position of intersection.

   Vector A:                        Vector B:

Position = Known                   Position = Known  
Direction= Known                   Direction= UNKNOWN
Magnitude= Known                   Magnitude= Known

To Find: Point of intersection.

Is it possible to find the point of intersection, with the given parameters? If so then how?

Application: I want to find the position where a player would be found based on the velocity he is moving, and shoot a bullet at him at the moment he would be found, taking into account the time taken for the bullet to reach that virtual target position.

Answer 1

Following on from the comments I'm going to take a leap here and answer your ultimate question directly.

Say the player is, at the initial time, at a point p and travelling with velocity v ; your gun is at position q and shoots a bullet in any direction at speed s :

The length of OP is vΔt and that of Q sΔt . The angle a is given by the dot product:

We can then use the cosine rule to solve for Δt :

Written in this form, we can easily see that it is a quadratic equation, and thus directly solve for Δt using the Quadratic formula :

There are a few cases we need to consider here:

• v < s : need to take the positive root only as otherwise we would get negative time.
• v > s and dot(PQ, s) < 0 : the bullet will never catch the player.
• v > s and dot(PQ, s) > 0 : take the negative root this time, as the positive root is for a backwards travelling player (longer time; this is also the case presented in the diagram above).

Having the correct value for Δt from above will then enable us to find the intersection point o , and therefore the intended direction d :

Note that d is not normalized. Also, this solution works for 3D too, unlike an approach with angles.

Answer 2

Let subscript 1 mark the player, and subscript 2 mark the AI:

• initial: position (x_i, y_i)
• angle: alpha_i
• speed: u_i

The positions as a function of time t are :

• player: (x_1 + u_1 * t * cos(alpha_1), y_1 + u_1 * t * sin(alpha_1))
• AI's bullet: (x_2 + u_2 * t * cos(alpha_2), y_2 + u_2 * t * sin(alpha_2))

You have 2 uknowns:

• t - the time of collision
• alpha_2 - the angle the AI should shoot

The collision happens when Xs and Ys match. ie:

• x_1 + u_1 * t * cos(alpha_1) = x_2 + u_2 * t * cos(alpha_2)
• y_1 + u_1 * t * sin(alpha_1) = y_2 + u_2 * t * sin(alpha_2)

So,

• alpha_2 = arcos( (x_1 + u_1 * t * cos(alpha_1) - x_2) / u_2 * t )

and also

• alpha_2 = arcsin( (y_1 + u_1 * t * sin(alpha_1) - y_2) / u_1 * t )

substitue your values and equate these to expressions of alpha_2 to obtain t, then you can substitue t in either expression to obtain the angle alpha_2.