Vector equation of a plane containing this line.

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Homework Help Overview

The problem involves finding the vector equation of a plane that contains a specific line given by the equation y=(3,2,1)+x(1,0,2) and is parallel to the z-axis. The context is within the subject area of vector geometry and equations of planes.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the substitution of values into the plane equation and question the appropriateness of using certain variables. There is an exploration of finding the normal vector to the plane and its relationship to the z-axis. Some participants express confusion about the correct interpretation of the plane's equation and the implications of the given line.

Discussion Status

The discussion is ongoing, with participants offering guidance on identifying the normal vector and clarifying the relationship between the line and the plane. There are multiple interpretations being explored, particularly regarding the coefficients in the plane equation and the implications of the plane being parallel to the z-axis.

Contextual Notes

Participants note the challenge of having multiple variables in the plane equation and the need for clarification on the relationship between the line's direction and the plane's normal vector. There is also mention of the specific point (3, 2, 1) lying on the plane, which is relevant to the discussion.

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Homework Statement



What is the vector equation of a plane that contains the line y=(3,2,1)+x(1,0,2) and is parallel to the z-axis?

Homework Equations



Equation of a plane: Ax + By + Cz + D = 0

The Attempt at a Solution



Ok so i substitute the values into the plane,
A(1)+B(0)+C(2)+D=0

But then I am left with 3 variables. Can you please offer some help?

Kind Regards,
 
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welcome to pf!

hi manmachine! welcome to pf! :smile:

sorry, but i don't understand what you're doing at all :redface:

the easiest way of finding the equation is probably first to find the normal to the plane …

what will that be? :wink:​

(and i think the question probably means the parametric equation v = (3,2,1) + t(1,0,2))
 


tiny-tim said:
hi manmachine! welcome to pf! :smile:

sorry, but i don't understand what you're doing at all :redface:

the easiest way of finding the equation is probably first to find the normal to the plane …

what will that be? :wink:​

(and i think the question probably means the parametric equation v = (3,2,1) + t(1,0,2))


Yes but I don't know what the plane is. :redface: I know that the normal is the coefficients if the plane. Please help.
 
manmachine said:

Homework Statement



What is the vector equation of a plane that contains the line y=(3,2,1)+x(1,0,2) and is parallel to the z-axis?

Homework Equations



Equation of a plane: Ax + By + Cz + D = 0

The Attempt at a Solution



Ok so i substitute the values into the plane,
A(1)+B(0)+C(2)+D=0
Why do you substitute those values? Those would be the components of a vector in the plane, not the coordinates of any point in the plane. (3, 2, 1) is a point in the plane. Setting x= 3, y= 2, z= 1, we have 3A+2B+ C+ D= 0.

Also, it is not a good idea to use "x" and "y" in that equation for the plane. In that vector equation "x" is a parameter, not the x- coordinate and "y" is a vector. Better to use, say, r= (3, 2, 1)+ t(1, 0, 2).

But then I am left with 3 variables. Can you please offer some help?

Kind Regards,
In that form, you can always divide through by any non=zero coefficient so you really have only two variables left. To eliminate one, use the fact that the plane is parallel to the z-axis. That means that the normal to the plane is also normal to the z-axis. That is, <0, 0, 1>.<A, B, C>= C= 0.
 
And yes, but how can I find the normal if don't have to vector equations IAXBI?
 
hi manmachine! :smile:
manmachine said:
Yes but I don't know what the plane is. :redface: .

yes, but you do know the directions of two lines that the normal is perpendicular to, don't you? :wink:
 
There is only one line Tim! I can't use cross product :(
 
z-axis? :smile:
 
oh fudge :P (bangs head on wall)
So it is 0,1,0 by cross product right?
 
  • #10
manmachine said:
oh fudge :P (bangs head on wall)

he he! :biggrin:
​
So it is 0,1,0 by cross product right?

uhh? :confused:
 
  • #11
Isn't that the normal to plane?
 
  • #12
(0,1,0) is the y-axis, and is normal to the xz-plane :confused:
 
  • #13
If I take the cross product of the z-axis (being 0,0,1) and our vector (1,0,2) this is what i get.
 
  • #14
oh i see! :rolleyes:

ok, now you know the normal to the plane, and you know at least one point on the plane,

so the vector equation of the plane is … ? :smile:​
 
  • #15
Am i right :)?

Take coefficents
0a+1b+0c+D=0
(0)(3)+(1)(1)+D=0
d=-1

This equation of our is plane is this:
b+-1=0 ?
 
  • #16
manmachine said:
Take coefficents
0a+1b+0c+D=0
(0)(3)+(1)(1)+D=0
d=-1

shouldn't there be a 2 in there somewhere? :confused:
This equation of our is plane is this:
b+-1=0 ?

what happened to x y and z ? :confused:

don't make it so complicated …

you know it's vertical, and parallel to the xz-plane, so it must be y = constant

ok, what is obviously that constant?? :smile:​
 
  • #17
I have no idea :p
 
  • #18
What do I do next . I am sorry I just don't get it
 
  • #19
This is very peculiar. You say that you know that any plane can be written as Ax+ By+ Cz+ D= 0. It has already been pointed out that you an divide through by any number, say, D, to reduce to three unknown constants, A'x+ B'y+ C'z+ 1= 0 (with A'
= A/D etc.). If the line given by the vector formula r= (3,2,1)+t(1,0,2) or parametric equations x= 3+ t, y= 2, z= 1+ 2t. In particular, taking t= 0, (3, 2, 1) is a point in the plane so we must have 3A'+ 2B'+ C'+ 1= 0.

Also, taking t= 1, (3+ 1, 2+ 0, 1+ 2)= (4, 2, 3) is also a point in the plane. so we must have 4A'+ 2B'+ 3C'+ 1= 0.

The last condition is that the plane is parallel to the z-axis. That means that its normal vector, <A, B, C> or <A', B', C'> is normal to <0, 0, 1>. Their dot product gives C= C'= 0 and so we have the two equations -3A'- 2B'- 1= 0 and 4A'+ 2B'+ 1= 0.

By the way, I find it simpler to use the fact that if a plane has normal vector <A, B, C> and contains point [itex](x_0, y_0, z_0)[/itex], then the plane has equation [itex]A(x- x_0)+ B(y- y_0)+ C(z- z_0)= 0[/itex].
 
  • #20
you know that the plane is y = constant

you also know that (3,2,1) lies on the plane

soooo, y must be … ? :smile:​
 
  • #21
3a'+2b'+c'+1=0?
 

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