Quick Vector intersection question

In summary: That's the point where the line intersects the yz-plane. In summary, a parametric equation for the line through P_1 (5,-2,4) and P_2 (2,6,1) is x = 5-3t, y = -2+8t, z = 4-3t. To find the points at which the line intersects each of the coordinate planes, set either x, y, or z to 0 and solve for t or s. In this case, setting x = 0 gives s = -2/3, and plugging this into the parametric equations gives the point (0, 34/3, -1) where the line intersects the y
  • #1
prace
102
0
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
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 each other 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
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
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
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
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.
 

Related to Quick Vector intersection question

What is a vector intersection and why is it important in science?

A vector intersection is a point where two or more vectors meet or cross each other. This is important in science because it allows us to analyze and calculate the relationship between different vectors and their directions.

How do you calculate the intersection point of two vectors?

To calculate the intersection point of two vectors, you can use the formula (x1,y1) + t(a,b) = (x2,y2) + s(c,d), where (x1,y1) and (x2,y2) are the starting and ending points of the two vectors, and (a,b) and (c,d) are the direction of the vectors. By solving for the values of t and s, you can find the coordinates of the intersection point.

What are some real-life applications of vector intersections?

Vector intersections have many real-life applications, such as in navigation systems, where the intersection of multiple vectors (e.g. GPS signals) can determine a precise location. They are also used in physics and engineering to calculate forces and velocities, and in computer graphics to render 3D images.

Can vector intersections occur in 3-dimensional space?

Yes, vector intersections can occur in 3-dimensional space. In fact, there can be multiple intersections between three or more vectors in 3D space, making it more complex to calculate the exact intersection point.

Are there any limitations or assumptions when using vector intersections?

One limitation when using vector intersections is that the vectors must be in the same plane in order to have an intersection point. Assumptions are also made about the nature of the vectors, such as their direction and magnitude, in order to accurately calculate the intersection point.

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