- #1

Darkmisc

- 210

- 28

- 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

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