How to find a point of intersection of two planes?

[Andrew Pierce]

Homework Statement

Two planes are given by the equations x + y + z = 1 for the plane P1 and x − y + z = 1 for the plane P2.
(or)
P1 : x + y + z = 1
P2 : x - y + z = 1

Q. The Question

Find the coordinates of a point of intersection of the planes P1 and P2

Homework Equations

Equation of a plane
Ax + By + Cz + D = 0

Parametric equations
x = xo + at
y = yo + bt
z = zo + ct

Possibly using cross product?
P1 x P2 ?

The Attempt at a Solution

Attempt 1:

So to start off I thought maybe finding the line of intersection would be the way to go about solving this problem, and then working from there to find some point.

Rewrite P1 and P2 to give the variable "x" on one side of the equation

P1 : x = 1 - y - z

P2 : x = 1 + y - z

Then setting both of the "x" variables equal to each other.

1 - y - z = 1 + y - z

Then I solve for "y"

y = 0

And now I'm lost.

Attempt 2:

Then I attempted to do what I normally did for trying to find any equations that intersect by setting the two equations equal to each other.

P1 = P2

x + y + z = x - y + z

And then I'm right back to where I started...

y = 0

Attempt 3:

Finally, I remembered a bit of information from a lecture from my professor about how using the cross product and setting a variable equal to zero was the way to go. Unfortunately, I do not have any detailed notes on the procedure and don't remember any more than that. P1 x P2 = < 2, 0, -2>

 

[tommyxu3]

The result you got is right, which means that every intersection of the two planes has the coordinate $(x,0,z).$ Then you can just use this to find out the complete solution of your problem.
Besides, the intersection of two planes will not a point, but nothing, a line or a plane.

 

[andrewkirk]

You've deduced that y must equal 0, so the intersection must lie in the x-z plane.
Now just substitute zero for y into one of the two original equations and you'll get an equation involving x and z that gives the line in the x-z plane that is the intersection.

 

[Andrew Pierce]

The result you got is right, which means that every intersection of the two planes has the coordinate $(x,0,z).$ Then you can just use this to find out the complete solution of your problem.
Besides, the intersection of two planes will not a point, but nothing, a line or a plane.

Right, I understand that but the problem I'm facing is, what do I do from there? I need to find a point on that line/plane I assume? But how?

 

[tommyxu3]

Now that you know the solution must present as $(x,0,z),$ then what you have to do is find the general types of $x$ and $z.$ From the remained relations:
$$x + y + z = 1,x - y + z = 1,$$
that is,
$$x+z=1.$$
This is the only constrict condition of this problem after you find $y=0$ is always true.

 

[Andrew Pierce]

Now that you know the solution must present as $(x,0,z),$ then what you have to do is find the general types of $x$ and $z.$ From the remained relations:
$$x + y + z = 1,x - y + z = 1,$$
that is,
$$x+z=1.$$
This is the only constrict condition of this problem after you find $y=0$ is always true.

So would a possible point for this line be (1, 0, 0)?
Or another, (0, 0, 1)?

Thanks so much for the help!

 

[tommyxu3]

So would a possible point for this line be (1, 0, 0)?
Or another, (0, 0, 1)?

Yes, they are all solutions, but mind what the problem requires. We need the general solution, that is, complete one. Knowing $x+z=1,$ try to use another parameter to present $x$ and $z$ and don't forget to give the range of your parameter, maybe $t.$

 

[Andrew Pierce]

Yes, they are all solutions, but mind what the problem requires. We need the general solution, that is, complete one. Knowing $x+z=1,$ try to use another parameter to present $x$ and $z$ and don't forget to give the range of your parameter, maybe $t.$

I do not want to sound stupid but I am unsure how to do that...
Maybe that is what has me stuck...
But to try and understand what it is do you simply substitute t for one of the variables and solve for each of the respective variables?

x = 1 - t
z = 1 - t

Is that it?

 

[tommyxu3]

It's close! But i don't know why you write down that. Knowing $x+z=1,$ if we set $x=t$ is a free parameter, then we can get $z=1-t.$ So the solution of this problem is $(x,y,z)=(t,0,1-t)~\forall t \in \mathbb{R}.$ Of course the choosing of the parameter is free depending on you and you will get another type looking different from this but in fact equivalent.

 

[tommyxu3]

For someone, they like to write the line in space like the type below:
$$\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c},$$
which here $(x_0,y_0,z_0)$ means a point on the line and $(a,b,c)$ is its normal vector.
For your problem, it should be: $x=1-z,~y=0.$

 

[Ray Vickson]

I do not want to sound stupid but I am unsure how to do that...
Maybe that is what has me stuck...
But to try and understand what it is do you simply substitute t for one of the variables and solve for each of the respective variables?

x = 1 - t
z = 1 - t

Is that it?

You obtained $1-y-z = 1+y-z$, so $y = 0$, necessarily. Now just go ahead and pick any value of $z$ you like, such as $z = - 6$ or $z = 147.2$ or anything else; all such choices will give you a point on the intersection.