Quick Vector intersection question

  • Thread starter prace
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  • #1
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So, I am to find a parametric equation for the line through [tex]P_1[/tex] and [tex]P_2[/tex] and then determine (if possible) the points at which the line intersects each of the coordinate planes.

[tex]P_1[/tex] (5,-2,4)
[tex]P_2[/tex] (2,6,1)

[tex]\vec{P_{1}P_{2}}[/tex] <-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!
 

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  • #2
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So I looked through my notes again to try and understand this, and I see that I found [tex]\vec{P_{2}P_{1}}[/tex] = <3,-8,3> and then found the parametric equation for the line from [tex]P_{2}[/tex] 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
HallsofIvy
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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
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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
radou
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prace said:
...
If a line passes through a coordinate plane, then at that point in time when it crosses that plane...

Actually, there's no time involved. But, if this makes it easier for you to think about parameters that way, then ok...
 
  • #6
HallsofIvy
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prace said:
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?
Yes, that's correct.
 

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