Find Planes in R^3 Intersecting xz-Plane: 3x + 2z = 5

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Homework Help Overview

The discussion revolves around identifying all planes in R^3 that intersect the xz-plane along the line defined by the equation 3x + 2z = 5. Participants are exploring the implications of the xz-plane being equivalent to y = 0 and how this affects the representation of planes in three-dimensional space.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants express confusion about how to start the problem and the significance of the y term in the plane's equation. Some suggest considering the direction vector of the line and the normal vector of the planes. Others question how to derive the coefficients for the planes and the implications of having infinite solutions.

Discussion Status

The discussion is ongoing, with participants providing insights into the nature of the problem and raising questions about the requirements for finding the planes. Some guidance has been offered regarding the relationship between the line and the planes, but there is still uncertainty about how to articulate the infinite nature of the solutions.

Contextual Notes

Participants note the challenge of interpreting the problem's request for "all planes" given the infinite possibilities that exist. There is also mention of needing to find additional equations that satisfy the conditions set by the intersection with the xz-plane.

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Homework Statement

Find all planes in R^3 whose intersection with the xz-plane is the lijne with equation 3x + 2z = 5

The Attempt at a Solution


Very confused here, not sure how to start it. the xz plane is another way of saying y = 0... which I'm guessing is why the equation doesn't have a y term. what is the best way to proceed?
 
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ggcheck said:
The xz plane is another way of saying y = 0... which I'm guessing is why the equation doesn't have a y term.

Your surmise is correct.

Think about the given line, which has the direction vector <3, 0, 2>. Pick a point on this line. Any plane you are interested in finding contains this line, so any vector perpendicular to its direction vector is the normal vector to one of those planes. What must those vectors look like? What would be the coefficients for the planes for which those are the normal vectors? What planes with those coefficients contain the point on 3x + 2z = 5 that you chose?
 
I'm still pretty confused
 
a point that satisfies 3x + 2z = 5 must also satisfy 0x + 1y + 0z = 0 and some other equation... so we need to find this other equation? how?
 
can you find at least one plane whose intersection with xz plane is that line?
 
uhhh
 
ggcheck said:

Homework Statement

Find all planes in R^3 whose intersection with the xz-plane is the lijne with equation 3x + 2z = 5

The Attempt at a Solution


Very confused here, not sure how to start it. the xz plane is another way of saying y = 0... which I'm guessing is why the equation doesn't have a y term. what is the best way to proceed?

Any plane can be written Ax+ By+ Cz= D for some A, B, C, D. you are correct that in the xz plane, y= 0. So saying that the plane intersects the xz plane in the line 3x+ 2z= 5, means that when y= 0, Ax+ Cz= D is the same as 3x+ 2z= 5. What are A, C, and D? What about B? (Remember that there are an infinite number of planes that intersect the xz plane in that line.)
 
Ax + Cz = 5?

cause A and c can be any numbers but D = 5?

ugh
 
no A=3 , C=2 , D=5 and B can be any number
 
  • #10
3x + 2z = 5 is the solution? It's that simple?
 
  • #11
3x + ny + 2z = 5

this is it, yes?
 
  • #12
I know this is pretty old, but I'm on the same problem and I'm stuck as well.

You say there's infinite planes that intersect with that line, but what exactly is the question asking for then?

It says "Find all planes in R^3", if there's infinite planes then what am I writing?

This is the last problem on my homework and I'm really confused about what it's asking for.
 

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