Solving Plane Equation 3x + 2y -z = 4

[Darkmisc]

Homework Statement

Find a Cartesian equation of the plane that is at right angles to the plane with the equation 3x + 2y - z =4 and goes through the points P(1, 2, 4) and Q(-1, 3, 2).

Relevant Equations

r⋅n=a⋅n

Hi everyone

One of the numbers in my attempt would hint that I have gotten something backwards in this question, but I can't see how.

For the plane 3x + 2y -z = 4, I've assumed the vector form is
r⋅(3i+2j -k) = 4. That is, (3i+2j -k) is the normal to the plane.

That being so, (3i+2j -k) will be parallel to plane that I am solving for. I'm going to use (3, 2, -1) as a point, M, on the plane relative to the origin.

PQ = (-2, 1, -2)
PM = (2, 0, -5)

The direction of PQxPM will be normal to the plane that I'm solving for.
PQxPM = n = -5i - 14j -2k

P⋅n and Q⋅n both give -41. The correct answer is -3x + 8y + 7z = 41.

I get -5x - 14y -2 = -41.

Does the -41 indicate that I'm on the right track, but with something flipped? Or have I taken the completely wrong approach?

Can someone show me how to solve this?Thanks

 

[Philip Koeck]

That is, (3i+2j -k) is the normal to the (given) plane.

That being so, (3i+2j -k) will be parallel to plane that I am solving for. I'm going to use (3, 2, -1) as a point, M, on the plane relative to the origin.

Why do you think M is in the plane you are looking for?

 

[Mark44]

Homework Statement:: Find a Cartesian equation of the plane that is at right angles to the plane with the equation 3x + 2y - z =4 and goes through the points P(1, 2, 4) and Q(-1, 3, 2).
Relevant Equations:: r⋅n=a⋅n

Hi everyone

One of the numbers in my attempt would hint that I have gotten something backwards in this question, but I can't see how.

View attachment 323596
For the plane 3x + 2y -z = 4, I've assumed the vector form is
r⋅(3i+2j -k) = 4. That is, (3i+2j -k) is the normal to the plane.

That being so, (3i+2j -k) will be parallel to plane that I am solving for. I'm going to use (3, 2, -1) as a point, M, on the plane relative to the origin.

PQ = (-2, 1, -2)
PM = (2, 0, -5)

(3, 2, -1) is a point on the given plane, but as the other responder said, you can't assume that's on the plane whose equation you need to find.

The direction of PQxPM will be normal to the plane that I'm solving for.
PQxPM = n = -5i - 14j -2k

That's the right idea, but you don't have the right normal to the new plane. For the plane you're trying to find, I get a normal of <3, -8, -7>. I have used this to determine the equation of the new plane, and have verified that the two given points satisfy the equation I found.

P⋅n and Q⋅n both give -41. The correct answer is -3x + 8y + 7z = 41.

I got an equivalent version of this equation -- namely, 3x - 8y - 7z = -41.

I get -5x - 14y -2 = -41.

Does the -41 indicate that I'm on the right track, but with something flipped? Or have I taken the completely wrong approach?

I think it's probably just a coincidence.

 

[haruspex]

(3, 2, 1) is a point on the given plane

I assume you meant (3, 2, -1), but that does not satisfy the equation of the given plane either.

 

[Mark44]

(3, 2, 1) is a point on the given plane

I assume you meant (3, 2, -1), but that does not satisfy the equation of the given plane either.

Yes, I meant (3, 2, -1). I've edited my earlier post to correct it. The OP came up with the point (3, 2, -1) in error.

Other than my typo above, which I didn't use in my calculations, the rest of my post is still accurate.

 

[WWGD]

Right. If planes are perpendicular, so are their normal vectors. Are your candidate vectors perpendicular?

 

[Mark44]

Right. If planes are perpendicular, so are their normal vectors. Are your candidate vectors perpendicular?

You're late to the party. I'm sure that the OP is aware of this, as he/she used the cross product to get a normal of the plane to be found.

 

[haruspex]

Yes, I meant (3, 2, -1). I've edited my earlier post to correct it. The OP came up with the point (3, 2, -1) in error.

Other than my typo above, which I didn't use in my calculations, the rest of my post is still accurate.

The given plane is $3x+2y-z=4$. (3, 2, -1) satisfies $3x+2y-z=14$, no?
Quite possibly that does not matter to your method.

 

[Mark44]

The given plane is $3x+2y-z=4$. (3, 2, -1) satisfies $3x+2y-z=14$, no?
Quite possibly that does not matter to your method.

No, it doesn't matter, although I misspoke when I said that the point (3, 2, -1) is on the given plane, or the point (3, 2, 1) for that matter. I'm glad you are catching these mistakes.