[_Lana_]

Hi guys

I need the equation of a plane that cuts the x-y plane through the line 2x+3y=6

and cuts the x-z plane with 2z+x=3


I'm so confused where to begin...does it have to do with reverse line intersection formula?

would be grateful for any hint

[ghostwalker]

I'm making this method up, so it may not be a standard one:

Bear in mind that the eqn of a plane or line is not unique, and any non-zero multiple of the eqn gives the same plane or line.

Now, we know the general equation of a plane is ax+by+cz+d=0 (or thereabouts).

When it cuts the x-y plane, what's the z value? Sub into the eqn of the plane and the line we get must be the given line, hence....

[aznkid66]

Line 1 can be put into the form 6-3y=2x.
Line 2 can be put into the form 3-2z=x.

x=3 (at intersection)
4x=12
x+2x+x=12

---

I don't know if that method actually returns the right answer, btw. Did you learn vectors yet? I would use those.

[ghostwalker]

(Original post by aznkid66)
Line 1 can be put into the form 6-3y=2x.
Line 2 can be put into the form 3-2z=x.

x=3 (at intersection)
4x=12
x+2x+x=12

---

I don't know if that method actually returns the right answer, btw. Did you learn vectors yet? I would use those.

Whilst I applaud the enthusiasm, I don't see the relevance to the OP's actual problem.

[aznkid66]

(Original post by ghostwalker)
Whilst I applaud the enthusiasm, I don't see the relevance to the OP's actual problem.

Sub in 2x in terms of y and x in terms of z and you get the equation x-3y-2z=3, which I later find to be wrong because you're setting x to one value while it has different values between the xy and xz plane. I'm was just wondering if you were required to find the equation of the line of intersection between the plane and the yz plane..

Vectors are obviously relevant. If two directional vectors of the plane are (3,2,0) and (2,0,1), then the normal is their cross, aka (2,-3,-4). The dot of the normal (2,-3,-4) and a point on the plane (3,0,0) is 6.

[_Lana_]

found it!!

it intersects with the zx plane (y=0 plane) at 2z + x = 3:
2x + 0 + kz = 6 has to be true.
But we also know that x + 2z = 3 when y = 0.
2x + kz = 6
x + 2z = 3
k = 4
Equation of the line is 2x + 3y + 4z = 6

[ghostwalker]

(Original post by _Lana_)
Equation of the line is 2x + 3y + 4z = 6

Yep.

Edit: I'm sure you meant plane there.

[ghostwalker]

(Original post by aznkid66)
Vectors are obviously relevant. If two directional vectors of the plane are (3,2,0) and (2,0,1)

Direction vectors would be (3,-2,0) and (2,0,-1).


I usually try and avoid vectors, if the question is specified in terms of co-ordinate geometry, and vice versa; though that's just me.

[the bear]

you can find two points on the line 2x + 3y = 6

for instance ( 0, 2, 0 ) and ( 3, 0, 0 )

and one point on the line 2z + x = 3

for instance ( 1, 0, 1 )

once you have 3 points on a plane it is straightforward to find the equation.