# Homework Help: Quick Vector intersection question

1. Oct 29, 2006

### prace

So, I am to find a parametric equation for the line through $$P_1$$ and $$P_2$$ and then determine (if possible) the points at which the line intersects each of the coordinate planes.

$$P_1$$ (5,-2,4)
$$P_2$$ (2,6,1)

$$\vec{P_{1}P_{2}}$$ <-3,8,-3>

x = 5-3t
y = -2+8t
z = 4-3t

So I got this far, but now I am not sure on what to do to find out if the line intersects the coordinate planes. I want to say that when the x, y, or z values are 0 in the parametric equations, is when the line intersects that plane, but I am not sure how to put it. If anyone can help me out there that would be great. Thanks!

2. Oct 29, 2006

### prace

So I looked through my notes again to try and understand this, and I see that I found $$\vec{P_{2}P_{1}}$$ = <3,-8,3> and then found the parametric equation for the line from $$P_{2}$$ which gave me:

x = 2+3s
y = 6-8s
z = 1+3s

Seeing that I then set each one equal to eachother and solved for t and s and found that they all came out to the same equation of 1 = t+s. Can anyone tell me what this means and if I am on the right track here?

3. Oct 30, 2006

### HallsofIvy

Just as, on the x-axis, y is 0 and on the y-axis, x= 0, on the coordinate planes, the remaining variable is 0.

On the xy-plane, z= 0. On the xz-plane, y= 0. On the yz-plane, x= 0.
If x= 2+ 3s= 0, then s= -2/3. What are y and z?

I have no idea what you mean by "set each one equal to each other"!

4. Oct 30, 2006

### prace

So, yeah... I don't know what I was doing either in my second post. That step was totally not needed. I mean, the two sets of parametric equations that I have here in Post #1 and Post #2 are the same, but in different directions, right? So I could use either one to find the points at which this line crosses the three planes.

If a line passes through a coordinate plane, then at that point in time when it crosses that plane, either the x, y, or z coordinate must be zero. So I can use this information to set either x, y, or z to zero, solve for t, or s, what have you, and find the remaining coordinates to construct the point at which the line passes through that coordinate plane. I think I got it.

So for this question, where you have solved for s = -2/3 with the x coordinate, the point where the line:

x = 2+3s
y = 6-8s
z = 1+3s

Passes through the yz plane is (0, 34/3, -1) How does that look?

5. Oct 30, 2006