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

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In summary, the student is very confused and needs help understanding what is required to find the solution to the equation 3x + 2z = 5. The student has identified that y = 0 in the xz plane, which means that any plane containing the line 3x+2z=5 also has the equation 0x + 1y + 0z = 0 as a normal vector. A, C, and D are the coefficients of the plane for which the normal vector is the positive direction, and B is the coefficient of the plane for which the normal vector is the negative direction. The plane with these coefficients intersects the xz plane at the point (3, 0, 2).
  • #1
ggcheck
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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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  • #2
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?
 
  • #3
I'm still pretty confused
 
  • #4
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?
 
  • #5
can you find at least one plane whose intersection with xz plane is that line?
 
  • #6
uhhh
 
  • #7
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.)
 
  • #8
Ax + Cz = 5?

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

ugh
 
  • #9
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.
 

What is the equation for finding planes in R^3 that intersect the xz-plane?

The equation for finding these planes is: 3x + 2z = 5.

What does the number 3 represent in the equation?

The number 3 represents the coefficient of the x variable, which is the slope of the plane in the x direction.

What does the number 2 represent in the equation?

The number 2 represents the coefficient of the z variable, which is the slope of the plane in the z direction.

What does the number 5 represent in the equation?

The number 5 represents the constant term, which is the y-intercept of the plane.

How many planes can intersect the xz-plane using this equation?

There can be an infinite number of planes that intersect the xz-plane using this equation, as there are infinite possible values for the x and z variables that satisfy the equation.

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