Find an equation of plane that contains the line

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Homework Help Overview

The problem involves finding an equation of a plane that contains a specified line and passes through a given point. The line is defined by the equation y = -2x + 3, and the plane is constrained to z = 5 while also passing through the point (2, -7, 23).

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss using different forms of the equation of a plane, with some suggesting the use of a normal vector approach while others prefer a non-vector method. There is exploration of how to derive the equation based on the line's slope and the point's coordinates.

Discussion Status

Some participants have provided guidance on using the general equation of a plane and have suggested methods for finding the necessary parameters. There is ongoing exploration of how to check the correctness of the derived equations and whether the approach aligns with the constraints of the problem.

Contextual Notes

Participants note that the class has just started teaching vectors, which may limit the methods available for solving the problem. There is also uncertainty about how to verify the derived equations against the original conditions of the problem.

Froster78
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I have looked at previous post that were on similar topics in hopes of not having to create a new post but I seem to be stuck still.

Homework Statement


Find an equation of the Plane that contains the line y=-2x+3 in the plane z=5, and goes through the point (2,-7,23)

Homework Equations


General equation of plane:
f(x,y,z)=mx+ny+c

The Attempt at a Solution


First I set up the equation of a plane using slopes from the equation of the line and values for the point it should pass through

23=-2(2)+n(-7)+3

then solved for y-slope

24=-7n

n=24/7

this gives me the function:

z=-2x-(24y/7)+3

I'm not sure my procedure was correct at all and I can't seem to figure out how to check this. If i plug in the values , I get the right z point, but this only tells me that I did my algebra correct (I hope :-!). I know I'm probably missing a major concept probably but some I cannot seem to make any progress on this problem.

Thanks for the help
 
Last edited:
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Froster78 said:
I have looked at previous post that were on similar topics in hopes of not having to create a new post but I seem to be stuck still.

Homework Statement


Find an equation of the Plane that contains the line y=-2x+3 in the plane z=5, and goes through the point (2,-7,23)


Homework Equations


General equation of plane:
f(x,y,z)=mx+ny+c



The Attempt at a Solution


First I set up the equation of a plane using slopes from the equation of the line and values for the point it should pass through

23=-2(2)+n(-7)+3

then solved for y-slope

24=-7n

n=24/7

this gives me the function:

z=-2x-(24y/7)+3

I'm not sure my procedure was correct at all and I can't seem to figure out how to check this. If i plug in the values , I get the right z point, but this only tells me that I did my algebra correct (I hope :-!). I know I'm probably missing a major concept probably but some I cannot seem to make any progress on this problem.

Thanks for the help

I don't think that's a very good expression for the general equation of a plane. Use ax+by+cz=k, with a,b,c,k all constants. Written in that form (a,b,c) is a normal vector to the plane. You can find a normal vector to the plane by finding two different vectors that are parallel to the plane and taking their cross product. Have you ever done it that way?
 
I find the easiest way to do it is to use the fact that any vector that lies in the plane will be normal to the plane normal (by definition)

If n is normal to the plane and if you have a point in the plane q=(u,v,w), now any and take any other point in the plane p=(x,y,z), the vector formed by the difference (p-q) between the points will be parallel to the plane.

The difference between every point outside the plane and p will give a vector that is not parallel to te hplane

the points in the plane p are exactly the ones given by
[tex](p-q) \bullet n = 0[/tex]
 
Yeah, I agree that vectors would be the best way to do this. Unfortunately we are supposed to use the "non vector" method. Although I am comfortable with vectors for the most part the class has just started teaching them. I think I see where I was wrong. I believe the equation that i need to use is:

z=m(x-x_0) + n(y-y_0) + z_0 that would give...

z = m(x-2) + n(y+7) + 23

I'm still not sure how to check the work... If I plug in 5 for z, should i get the equation of the line after solving for y?
 
Last edited:
Froster78 said:
Yeah, I agree that vectors would be the best way to do this. Unfortunately we are supposed to use the "non vector" method. Although I am comfortable with vectors for the most part the class has just started teaching them. I think I see where I was wrong. I believe the equation that i need to use is:

z=m(x-x_0) + n(y-y_0) + z_0 that would give...

z = m(x-2) + n(y+7) + 23

I'm still not sure how to check the work... If I plug in 5 for z, should i get the equation of the line after solving for y?

Ok, I think that form will work. But you still need to find m and n. Plug in a couple of points on the line y=-2x+3 in the plane z=5 and try to solve for m and n.
 

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