1. ## 3D Vector Calculation

I'm a student who is currently in a calculus course, and I would like to work a bit further ahead for understanding. I have thought of a good scenario to help illustrate what I would like to learn, so if anybody could explain how the calculations would work, I would greatly appreciate it.

In a playing field, two people are playing frisbee. Joe has the frisbee, and he is located at coordinates (-50, -80), and the height of the frisbee he is holding in his hand (which is currently right in front of his face) is +100 (z). He is currently looking at a tree out in the distance located at (30, 40), at the same height as his frisbee (+100).

His friend, Bob, is located at (20, -150). Joe wants to throw the frisbee into Bob's hands, which are currently outstretched into the air at about +200 (z).

How would I calculate the vector for the new direction Joe would need to face in order for the frisbee to be aimed exactly at where Bob's hands are? (Assuming that Joe's frisbee is still in front of his face.)

As well, if anybody would like to help further, how would I take velocity into consideration, if Bob is running a certain direction, and Joe wants to throw it ahead of him in order for him to catch it when the frisbee reaches that point? (Joe won't actually throw the frisbee yet, he just needs to be looking at the right location where Bob is predicted to be at.)

And how would that change if Bob were to jump? (Going up and then back down.)

Thank you all very much! I love math, and I really love learning things beforehand ^^.

2. He will need, of course, to face toward Bob. That is, he would have to face from his own position, (-50, -80, 100), toward Bobs position, (20, -50, 200). That vector is <20-(-50), -50-(-80), 100- 200>= <70, 30, -100>.

But are you sure that is the question? Since they give his original orientation as facing from (-50, -80, 100) to the tree, at (30, 40 100), his current "facing vector" is <30-(-50), 40-(-80), 100- 100>= <80, 120, 0>.

To find the angle through which he must turn, you can use the dot product and the fact that the angle, $\displaystyle \theta$, between vector u and v is given by $\displaystyle cos(\theta)= \frac{u\cdot v}{|u||v|}$.

3. Originally Posted by HallsofIvy
He will need, of course, to face toward Bob. That is, he would have to face from his own position, (-50, -80, 100), toward Bobs position, (20, -50, 200). That vector is <20-(-50), -50-(-80), 100- 200>= <70, 30, -100>.

But are you sure that is the question? Since they give his original orientation as facing from (-50, -80, 100) to the tree, at (30, 40 100), his current "facing vector" is <30-(-50), 40-(-80), 100- 100>= <80, 120, 0>.

To find the angle through which he must turn, you can use the dot product and the fact that the angle, $\displaystyle \theta$, between vector u and v is given by $\displaystyle cos(\theta)= \frac{u\cdot v}{|u||v|}$.
Yes, I am trying to calculate the angle that Joe must turn in order to be looking at Bob's hands.