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Find point of intersection of a line and a plane

  • Thread starter Jatt
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  • #1
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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.
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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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
7
0
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:

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