Finding the line of intersection of 2 planes

In summary, the planes 5x-2y-2z = -1 and x-4y+2z = 25 are not parallel and must intersect along a line that is common to both of them. The vector parametric equation for this line is r= (-4, 0, 29/2)+ t(1, 1, -3/2).
  • #1
ih8calc
4
0

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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  • #2
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].
 
  • #3
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.
 
  • #4
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:
  • #5
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...
 

Related to Finding the line of intersection of 2 planes

1. What is the line of intersection of 2 planes?

The line of intersection of 2 planes is the path where the 2 planes intersect in 3-dimensional space. It is the set of points that lie on both planes at the same time.

2. How do you find the line of intersection of 2 planes?

To find the line of intersection of 2 planes, you need to first find the equations of both planes. Then, you can use algebraic methods such as substitution or elimination to solve for the variables and find the values that satisfy both equations. The resulting solution will be the equation of the line of intersection.

3. Can the line of intersection of 2 planes be parallel?

Yes, it is possible for the line of intersection of 2 planes to be parallel. This occurs when the planes are parallel to each other or when they are coincident (overlapping each other).

4. How many solutions are there for finding the line of intersection of 2 planes?

There can be three possible scenarios when finding the line of intersection of 2 planes: 1) the planes intersect at a single point, resulting in one solution, 2) the planes are parallel, resulting in no solution, or 3) the planes are coincident, resulting in infinitely many solutions.

5. In what situations would you need to find the line of intersection of 2 planes?

Finding the line of intersection of 2 planes is useful in various fields such as engineering, physics, and geometry. It can be used to determine the point of intersection between two objects, to calculate angles and distances in 3-dimensional space, or to find the common boundary between two regions.

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