Point of intersection of a line and plane. No points in plane given. Wheres SUPERMAN?

  • Thread starter Hellken
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  • #1
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Homework Statement



I need to locate the coordinates if a point of intersection x0,y0,z0 of a plane with equation 2x+y-z=0 and a line that is perpendicular to that plane and passes through a point G(2,1,0).

Homework Equations

and

I understand that this is a normal line and a plane so the dot product: {Normal Line}`{P} = 0.

The dot product equals the equation of the plane(right or Is that inaccurate?): {Normal Line}'{P}= a(x-x0)+b(y-y0)+c(z-z0) = 0

The Attempt at a Solution



(I have been at this for hours so I have a clump of scratch paper so this is my recent desperate attempt to hack it, im tired and desperate).

My plan: Im given the equation of the plane and a point outside the plane where the line passes through. So essentially I solve for the point of intersection and another point in the plane p(x,y,z) ?

{Normal Line}'{P}= a(x-x0)+b(y-y0)+c(z-z0) = 0
(a) So, I have a normal line from (2,1,0) to pt. of intersection (x0,y0,z0) and get the vector eqn of that ?

Important: 1) When solving is the solution of the normal line and plane a point of intersection just as the solution of two intersecting lines ?

2) Do I necessarily have to solve for another point in the plane perpendicular to the normal line the point of intersection? I have been wrecking this for some time but nothing adds up in my clumps of scratch paper.

3) Do I need a parametric representation of the lines?
 

Answers and Replies

  • #2
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Oh shoot. This isnt a homework but I have an upcoming exam in less than 24 hours and some solutions and answer would be nice.

Another question but the normal vector perpendicular to the plane is <2,1,-1> right? Where to from here?

================================================================================================
I have a test in a few hours and I would appreciate it if someone could spell it out and give me a good rundown. This question in my textbook is paradoxical but solvable and my prof loves giving strange questions like this. I would really appreciate it if someone could brief me ASAP. ================================================================================================
 
  • #3
LCKurtz
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Use the normal to the plane as the direction vector for the line. Write parametric equations for that line expressing x, y, z in terms of t. Plug those x,y,z's into the plane to get the value of t and plug that value of t into the line to get the intersection coordinates.
 
  • #4
HallsofIvy
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Homework Statement



I need to locate the coordinates if a point of intersection x0,y0,z0 of a plane with equation 2x+y-z=0 and a line that is perpendicular to that plane and passes through a point G(2,1,0).
A vector perpendicular to the plane is 2x+ y- z= 0 is <2, 1, -1> and you want a line perpendicular to the line through (2, 1, 0) so the line is x= 2t+ 2, y= t+ 1, z= -t.

Plug those into 2x+ y- z= 0 and solve for t.

Homework Equations

and

I understand that this is a normal line and a plane so the dot product: {Normal Line}`{P} = 0.

The dot product equals the equation of the plane(right or Is that inaccurate?): {Normal Line}'{P}= a(x-x0)+b(y-y0)+c(z-z0) = 0

The Attempt at a Solution



(I have been at this for hours so I have a clump of scratch paper so this is my recent desperate attempt to hack it, im tired and desperate).

My plan: Im given the equation of the plane and a point outside the plane where the line passes through. So essentially I solve for the point of intersection and another point in the plane p(x,y,z) ?

{Normal Line}'{P}= a(x-x0)+b(y-y0)+c(z-z0) = 0
(a) So, I have a normal line from (2,1,0) to pt. of intersection (x0,y0,z0) and get the vector eqn of that ?

Important: 1) When solving is the solution of the normal line and plane a point of intersection just as the solution of two intersecting lines ?

2) Do I necessarily have to solve for another point in the plane perpendicular to the normal line the point of intersection? I have been wrecking this for some time but nothing adds up in my clumps of scratch paper.

3) Do I need a parametric representation of the lines?
 
  • #5
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LCKurtz & Halls of Ivy thank you for your timely replies. I understand where the both of you are headed but thats where the paradox in this problem is. With no knowledge of any other points on the plane 2x +y-z=0, I am unable to properly solve for "t" properly because I have "leftover" x's, y's and z's. So while the usual procedure is :

1) Find a set of parametric eqns to represent the line (In this case I chooose the normal line because we are looking for the intersection).
2) Solve for "t"
3) Once "t" is obtained plug it in.

I am unable to get past stage 2 due to a flaw in stage one simply because I lack knowledge of other points (x,y,z) on the plane that are perpendicular to the normal line.

Are there any other methods of finding a set of parametric eqns to describe the normal line then?
 
Last edited:
  • #6
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Hellken said:
With no knowledge of any other points on the plane 2x +y-z=0,...

Since you have the formula for the plane, you can generate as many points on it as you want, simply by choosing values for any two variables, then solving for the third.

Hellken said:
I lack knowledge of other points (x,y,z) on the plane that are perpendicular to the normal line.
This doesn't make much sense to me. A point isn't perpendicular to a line.

Maybe I'm missing something here...
 
  • #7
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==========================================================================================================
EDIT: Current solution:

Dot product: N*PP0=<a,b,c>*<x-x0,y-y0,z-z0> = 0

--------------------------
NB: I have a point (2,1,0) on my normal line N that intersects the plane at x0,y0,z0 .
Lets see where this goes...


a = 2 - x0
b = 1 - y0
c = -z0

-------------------------
The eqn of the plane:
a(x-x0) + b (y-y0) + c(z-z0) = 0

Substituting a, b, and c:

(2 - x0)(x-x0) + (1 - y0)(y-y0) + (-z0)(z-z0) = 0

----------------------------

Back to finding a parametric eqn of the normal line N:

N = r(t) = <2,1,0> + t<x0 - 2, y0 - 1, z0>
N = r(t) = <2 + t(x0) - 2t, (t)y0- t + 1, (t)z0>

At this point I am in a sea of "leftover" variables and unable to isolate "t".
 
Last edited:
  • #8
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Since you have the formula for the plane, you can generate as many points on it as you want, simply by choosing values for any two variables, then solving for the third.


This doesn't make much sense to me. A point isn't perpendicular to a line.

Maybe I'm missing something here...

Im pretty new here so does this forum have a drawing tool? I mean that there is a line Normal to the plane. It intersects at a certain specific point in the plane (What the problem in my textbook demands). The normal line is perpendicular to the plane so it is perpendicular to a line on the plane.

You are right in saying that I can generate as many points as I want. True but I am stumped here since I cannot "feel" my way around the plane. I am only given one point (2,1,0) and that the one point I am given is "attached" to the plane by the normal line that intersects it. I'm sure I'm missing something elementary here that's all. Either that or Ive complicated the problem further.

Boy I need a vacation soon. Its hard to think straight at a time like this.
 
Last edited:
  • #9
13
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UPDATE: It's a bird. It's a plane. No its Captain Obvious!

People forget the previous equations and forget it all. I got it and it was really simple. I forgot my "v"! I knew something was missing! It was a simple elementary mistake.

I'll update on my solution (when I get back) if anyone else insists. Thanks for all your help to those who replied back!

I need more sleep...
 

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