Equation of plane

[PcMogu]

Find an equation of the plane through the line of intersection of the planes x - z = 1 (lets call this A) and y - 2z = 3 (lets call this B) and is perpendicular to plane x + y - 2z = 1 (lets call this C).
I am not sure where to start. I think that I can do the cross product from the intersection of the two planes. A is (1,0,-1) and B is (0,1,-2) cross those to get <1,2,1> and then maybe cross that with the normal of C which is (1,1,-2) to get <-5,3,-1> meaning maybe that the eq is something like -5 ( x - ) + 3 ( y - ) -1 ( z - ) = 0. If that is right, where do I get the points? Thanks.

 

[galactus]

The two given planes can be expressed parametrically:

\(\displaystyle x-z-1=0\)

\(\displaystyle y-2z-3=0\)

These can be solved by subbing or eliminating to get:

\(\displaystyle x=t, \;\ y=1+2t, \;\ z=t-1\)

Now, v=[1,2,1] is parallel to the line and n=[1,1,-2] is normal to the given planes so

\(\displaystyle \boxed{v\times n=[-5,3,-1]}\) is normal to the plane we want.

 

[PcMogu]

The two given planes can be expressed parametrically:

\(\displaystyle x-z-1=0\)

\(\displaystyle y-2z-3=0\)

These can be solved by subbing or eliminating to get:

\(\displaystyle x=t, \;\ y=1+2t, \;\ z=-t+1\)

Now, v=[1,2,-1] is parallel to the line and n=[1,1,-2] is normal

Alright I didn't quite understand your first step. I tried getting the normals off the two lines, crossing them \(\displaystyle (1,0,-1) x (0,1,-2) = <1,2,1>\). You got <1,2,-1>, why the -1? What did I do wrong?

 

[JuicyBurger]

The two given planes can be expressed parametrically:

\(\displaystyle x-z-1=0\)

\(\displaystyle y-2z-3=0\)

These can be solved by subbing or eliminating to get:

\(\displaystyle x=t, \;\ y=1+2t, \;\ z=-t+1\)

Now, v=[1,2,-1] is parallel to the line and n=[1,1,-2] is normal

Alright I didn't quite understand your first step. I tried getting the normals off the two lines, crossing them \(\displaystyle (1,0,-1) x (0,1,-2) = <1,2,1>\). You got <1,2,-1>, why the -1? What did I do wrong?

In order to express the planes parametrically you need to start with x = t, which is standard. From here you can use the first equationx - z - 1 = 0 to solve for z, plugging in x = t.

This should give you:
x = t
z = t-1 (not z = -t + 1)

Now you can substitue this into the next equation, y - 2z - 3 = 0 to get y = 2t + 1.

 

[JuicyBurger]

So yes, your vector of \(\displaystyle <1,2,1>\) is correct.

 

[Aladdin]

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The equation of the plane is : -5i + 3j - k .

 

[galactus]

You got <1,2,-1>,

Sorry, that was a typo. I fixed my foolish error. Thanks. Goes to show you, one little negative sugn misplaced and it's all kaput.