Find point of intersection of a line and a plane

In summary, the point of intersection between the line and plane is P(2,6,0). The equations were solved by eliminating the variables s and t and solving for k, which was then substituted into the equations of the line to find the point of intersection. The point was verified by checking its coordinates in the original equations.
  • #1
Jatt
7
0

Homework Statement


Find the point of intersection of the line and the plane.
Line: x=4+k
y=2-2k
z=6+3k
Plane: x=-1-s+2t
y=1-s+4t
z=2+3s+t



Homework Equations


None.



The Attempt at a Solution


So I'm not that well informed with how these lines and planes behave... With that being said my attempt may be off.

What really confuses me is the parametric form of the equation of the plane. I've never seen it in this form before. I'm used to dealing with the form Ax+By+Cz+D=0. So I have no idea of what I'm suppose to do with the two variables s and t...
I tried equating the equations of the line with the equations of the plane and just got stuck with no clue of what to do.

4+k = -1-s+2t ---- Equation 1
2-2k=1-s+4t ----- Eq 2
6+3k=2+3s+t ------ Eq 3

-2 + Eq 2
---------------
-8-2k=2+2s-4t
+ 2-2k=1-s+4t
---------------
-6-4k=3+s
-------------
With no idea of what I was actually doing I decided to give up here and look through my school textbook which really didn't explain much. I also couldn't find any similar questions through googizing. I hope you guys can help me understand how to solve this question.
Thanks.
 
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  • #2
Jatt said:

Homework Statement


Find the point of intersection of the line and the plane.
Line: x=4+k
y=2-2k
z=6+3k
Plane: x=-1-s+2t
y=1-s+4t
z=2+3s+t



Homework Equations


None.



The Attempt at a Solution


So I'm not that well informed with how these lines and planes behave... With that being said my attempt may be off.

What really confuses me is the parametric form of the equation of the plane. I've never seen it in this form before. I'm used to dealing with the form Ax+By+Cz+D=0. So I have no idea of what I'm suppose to do with the two variables s and t...
I tried equating the equations of the line with the equations of the plane and just got stuck with no clue of what to do.
Exactly right! You now have 3 equations to solve for the 3 unknowns. Although to answer the question, you don't need to find all 3. Probably the simplest thing to do is eliminate s and t so you can solve for k. Then put that value of k back into the equations of the line. The x, y, z they give will be the point of intersection.

4+k = -1-s+2t ---- Equation 1
2-2k=1-s+4t ----- Eq 2
6+3k=2+3s+t ------ Eq 3

-2 + Eq 2
You mean "-2(Eq 1)+ Eq 2" don't you?
---------------
-8-2k=2+2s-4t
+ 2-2k=1-s+4t
---------------
-6-4k=3+s
-------------
With no idea of what I was actually doing I decided to give up here and look through my school textbook which really didn't explain much. I also couldn't find any similar questions through googizing. I hope you guys can help me understand how to solve this question.
Thanks.
So far you have eliminated t from equations 1 and 2. Now eliminate t using equations 1 and 3 or 2 and 3: for example, -2(Eq 3)+ Eq 1 will also eliminate t, leaving two equations in k and s. Combine the two equations so as to eliminate s and you will be left with one equation to solve for k.
 
  • #3
Yea your right that's what i meant -2(Eq 1) + Eq 2.
I was doing it right all along? Cool. I understand what I have to do next it seems much more less confusing than before. Thanks a lot now I can finally finish this question off. :D

Just finished I got the point of intersection between the line and plane to be P(2,6,0). I also checked the answers I got for my variables by substituting them into the first equations which I equated. I found both sides for all 3 equations equal its opposite side. Which means the line and plane intersect each other. So my point should be correct.
Thanks!
 
Last edited:

1. What is the formula for finding the point of intersection of a line and a plane?

The formula for finding the point of intersection of a line and a plane is to first find the intersection of the line and the plane's normal vector. This can be done by solving the system of equations created by the line's parametric equations and the plane's equation in standard form. Once the values for the parameters are found, they can be substituted into the line's equations to find the coordinates of the point of intersection.

2. Can a line and a plane have more than one point of intersection?

Yes, a line and a plane can have more than one point of intersection. This can occur when the line is parallel to the plane, in which case it will intersect the plane at every point along its length. It can also occur when the line lies within the plane, as it will intersect the plane at every point along its length as well.

3. How do you know if a line and a plane are parallel?

If the line's direction vector is perpendicular to the plane's normal vector, then the line and the plane are parallel. This means that the line will never intersect the plane, unless it lies entirely within the plane.

4. Is it possible for a line and a plane to not have a point of intersection?

Yes, it is possible for a line and a plane to not have a point of intersection. This can occur when the line is parallel to the plane, or when the line lies in a different plane that does not intersect the given plane.

5. How can finding the point of intersection of a line and a plane be useful in real life?

Finding the point of intersection of a line and a plane can be useful in many real-life scenarios, such as in engineering, architecture, and physics. For example, in architecture, this calculation can help determine where a roof or a beam will intersect with a wall or another structure. In physics, it can be used to find the point of impact of a projectile on a surface. Additionally, in computer graphics, this calculation is used to create realistic 3D images and animations.

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