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

• Darkmisc
In summary, the person is looking for the equation of a plane that is perpendicular to the one with the equation 3x + 2y - z = 4 and goes through points P(1, 2, 4) and Q(-1, 3, 2). The person has found a new plane that has the equation 3x - 8y - 7z = -41.
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

Darkmisc said:
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?

Darkmisc
Darkmisc said:
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.
Darkmisc said:
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.
Darkmisc said:
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.
Darkmisc said:
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.

Last edited:
Darkmisc
Mark44 said:
(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 said:
(3, 2, 1) is a point on the given plane
haruspex said:
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.

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

WWGD said:
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.

SammyS and WWGD
Mark44 said:
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.

haruspex said:
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.

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

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## What is the general form of a plane equation?

The general form of a plane equation in three-dimensional space is Ax + By + Cz = D, where A, B, C, and D are constants. In this specific case, the equation is 3x + 2y - z = 4.

## How do you find the normal vector of the plane 3x + 2y - z = 4?

The normal vector of the plane can be found from the coefficients of x, y, and z in the plane equation. For the equation 3x + 2y - z = 4, the normal vector is (3, 2, -1).

## How can you determine if a point lies on the plane 3x + 2y - z = 4?

To determine if a point (x, y, z) lies on the plane, substitute the coordinates of the point into the equation 3x + 2y - z = 4. If the equation holds true, then the point lies on the plane. For example, the point (1, 1, 1) lies on the plane because 3(1) + 2(1) - 1 = 4.

## What is the distance from a point to the plane 3x + 2y - z = 4?

The distance from a point (x0, y0, z0) to the plane Ax + By + Cz = D is given by the formula |Ax0 + By0 + Cz0 - D| / sqrt(A^2 + B^2 + C^2). For the plane 3x + 2y - z = 4 and a point (x0, y0, z0), the distance is |3x0 + 2y0 - z0 - 4| / sqrt(3^2 + 2^2 + (-1)^2).

## How do you find the intersection line of the plane 3x + 2y - z = 4 with another plane?

To find the intersection line of two planes, you need to solve their equations simultaneously. For example, if the second plane is given by 2x - y + z = 1, you can solve the system of linear equations:1. 3x + 2y - z = 42. 2x - y + z = 1By solving these equations, you can find the parametric equations of the line of intersection.



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