Finding the line of intersection of 2 planes

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Homework Help Overview

The problem involves finding the line of intersection between two planes defined by the equations 5x-2y-2z = -1 and x-4y+2z = 25. Participants note that since the planes are not parallel, they must intersect along a line.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss adding the plane equations and manipulating them to express variables in terms of one another. There are attempts to derive parametric equations for the line of intersection. Some participants express uncertainty about their algebraic manipulations and seek confirmation of their reasoning.

Discussion Status

Several participants have provided different approaches to finding the line of intersection, including using normal vectors and cross products. There is ongoing exploration of how to derive a specific point on the line and the corresponding parametric equations. No consensus has been reached, but multiple lines of reasoning are being examined.

Contextual Notes

Participants are working under the constraints of homework guidelines, which may limit the extent of assistance they can provide to one another. There is also a noted confusion regarding the manipulation of equations and the selection of parameters.

ih8calc
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Homework Statement


The planes 5x-2y-2z = -1 and x-4y+2z = 25 are not parallel, so they must intersect along a line that is common to both of them. The vector parametric equation for this line is ...


The Attempt at a Solution



Ugh... this is a doozie...

So I started the problem by adding both plane equations

5x - 2y - 2z = -1 (1)
x - 4y + 2z = 25 (2)

I get

6x-6y=24
6x = 6y-24

now here is where I run a little brain snag...

x = y -4 (I'm not sure if this is correct... shouldn't it be x = (6y-24)/6?) anyway carry on

so then substituting x into the second equation up there...I gets

y-4-4y+2z=25
-3y + 2z = 29
2z = 29 - 3y
z = 29/2 - 3y/2

I have, hopefully, done this correctly and got x= y - 4 and z =29/2-3y/2
t = y
I need to find y and the parametric equation for it

so far i got (- 4, *, 29/2)+t(1, *,-3/2)

please halp...
 
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ih8calc said:

Homework Statement


The planes 5x-2y-2z = -1 and x-4y+2z = 25 are not parallel, so they must intersect along a line that is common to both of them. The vector parametric equation for this line is ...


The Attempt at a Solution



Ugh... this is a doozie...

So I started the problem by adding both plane equations

5x - 2y - 2z = -1 (1)
x - 4y + 2z = 25 (2)

I get

6x-6y=24
6x = 6y-24

now here is where I run a little brain snag...

x = y -4 (I'm not sure if this is correct... shouldn't it be x = (6y-24)/6?)
x = y - 4 is equivalent to x = (6y - 24)/6.

ih8calc said:
anyway carry on

so then substituting x into the second equation up there...I gets

y-4-4y+2z=25
-3y + 2z = 29
2z = 29 - 3y
z = 29/2 - 3y/2

I have, hopefully, done this correctly and got x= y - 4 and z =29/2-3y/2
t = y
I need to find y and the parametric equation for it

so far i got (- 4, *, 29/2)+t(1, *,-3/2)

please halp...

You haven't done enough that I can check your work, so here's another approach.
  1. Find a normal to plane 1. You can do this by inspection.
  2. Find a normal to plane 2. ditto
  3. Calculate the cross product of these normals. This will give you a vector that is perpendicular to both normals, and that will have the same direction as line you want. Let's say that the coordinates of this vector are (A, B, C)
  4. Find a point that is common to both planes, by using the equation you found, x = y - 4. In this equation, z and y are arbitrary. Let's say that the point is [tex](x_0, y_0, z_0)[/tex].
The equation of your line will be [tex](x, y, z) = (x_0, y_0, z_0) + t(A, B, C)[/tex].
You should be able to find values for [tex]x_0, y_0, z_0, A, B, and C[/tex].
 
n1 = <5, -2, -2>
n2 = <1, -4, 2>
n1 x n2 = <-12,-12,-18>

mkay... so now

(x, y, z) = (x0, y0, z0) + t(-12, -12, -18)

and this point
x = y -4 is used to find (x0,y0,z0)...

I'm stuck, yuck.
 
ih8calc said:
x = y -4 is used to find (x0,y0,z0)...

I'm stuck, yuck.

Pick any value for y (or x) and then find the corresponding value for x (or y). For example, put y=0 and find x.

the put those values into either equations for the planes and get z. You will now have a point on the line that you want to get.

EDIT: If you wanted to, you could have just let y=t and just find x and z in terms of t and you'd get that vector line.
 
Last edited:
ih8calc said:

Homework Statement


The planes 5x-2y-2z = -1 and x-4y+2z = 25 are not parallel, so they must intersect along a line that is common to both of them. The vector parametric equation for this line is ...


The Attempt at a Solution



Ugh... this is a doozie...

So I started the problem by adding both plane equations

5x - 2y - 2z = -1 (1)
x - 4y + 2z = 25 (2)

I get

6x-6y=24
6x = 6y-24

now here is where I run a little brain snag...

x = y -4 (I'm not sure if this is correct... shouldn't it be x = (6y-24)/6?)
Yes, and (6y- 24)/6= (6y/6)- (24/6)= y- 4

At this point, I think I would let y itself be the parameter: y= t so x= t- 4. Now put those back into x- 4y+ 2z= 25: t-4- 4t+ 2z= 25 and solve for z. 2z= 25+ 3t+ 4= 29+ 3t so z= 29/2+ (3/2)t.
Parametric equations for the line are x= t-4, y= t, and z= 29/2- (3/2)t.
The vector equation would be r= (-4, 0, 29/2)+ t(1, 1, -3/2).

anyway carry on

so then substituting x into the second equation up there...I gets

y-4-4y+2z=25
-3y + 2z = 29
2z = 29 - 3y
z = 29/2 - 3y/2

I have, hopefully, done this correctly and got x= y - 4 and z =29/2-3y/2
t = y
I need to find y and the parametric equation for it

so far i got (- 4, *, 29/2)+t(1, *,-3/2)

please halp...
 

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