Find an equation of plane that contains the line

[Froster78]

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

 

[Dick]

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?

 

[lanedance]

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
(p-q) \bullet n = 0

 

[Froster78]

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?

 

[Dick]

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.