Finding the Equation of a Plane Perpendicular to a Given Line and Plane

  • Thread starter BraedenP
  • Start date
  • Tags
    Plane
In summary, the equation of the plane that contains the given line and is perpendicular to the plane 2x-4y+2z=9 is <2, -4, 2> multiplied by a scalar value, which can be determined by using the point specification of the given line and plugging it into the plane equation.
  • #1
BraedenP
96
0

Homework Statement


Find the equation of the plane that contains the line [itex]x=-1+3t, y=5+2t, z=2-t[/itex] and is perpendicular to the plane [itex]2x-4y+2z=9[/itex]

Homework Equations



Equation of a plane:
[tex]a(x-x_o)+b(y-y_o)+c(z-z_o)=0[/tex]

The Attempt at a Solution


I was thinking that I could simply find an orthogonal normal to the normal of the specified plane. That would give me the normal of my new plane.

I found such a normal to be (3, 1, -1).

Then I'd simply take a the direction of the line (3, 2, -1), and plop it into the plane equation. This is where I get stuck; I can't figure out how to get the point specification into that equation. Is it simply:

[tex]3(3)x+1(2)y-1(-1)z = 9x+2y+z = 0[/tex]

It seems too easy for me. What am I doing wrong?
 
Last edited:
Physics news on Phys.org
  • #2
BraedenP said:

Homework Statement


Find the equation of the plane that contains the line [itex]x=-1+3t, y=5+2t, z=2-t[/itex] and is perpendicular to the plane [itex]2x-4y+2z=9[/itex]


Homework Equations



Equation of a plane:
[tex]a(x-x_o)+b(y-y_o)+c(z-z_o)=0[/tex]

The Attempt at a Solution


I was thinking that I could simply find an orthogonal normal to the normal of the specified plane. That would give me the normal of my new plane.

I found such a normal to be (3, 1, -1).
This is one vector that is perpendicular to the normal of the plane you are to find. The problem is that there are an infinite number of vectors that are perpendicular to that plane.
BraedenP said:
Then I'd simply take a the direction of the line (3, 2, -1), and plop it into the plane equation. This is where I get stuck; I can't figure out how to get the point specification into that equation. Is it simply:

[tex]3(3)x+1(2)y-1(-1)z = 9x+2y+z = 0[/tex]

It seems too easy for me. What am I doing wrong?
 
  • #3
Mark44 said:
This is one vector that is perpendicular to the normal of the plane you are to find. The problem is that there are an infinite number of vectors that are perpendicular to that plane.

There are, but all the rest of them would simply be scalar multiples of that one, right? Therefore, it doesn't matter which multiple is used in the equation; it'll work either way.
 
  • #4
No, not at all. All of the vectors that are perpendicular to <2, -4, 2> would lie in the same plane, but they point in all different directions.
 

1. What is the equation of a plane?

The equation of a plane is a mathematical representation of a flat, two-dimensional surface in three-dimensional space. It is typically written in the form of Ax + By + Cz + D = 0, where A, B, and C are the coefficients of the x, y, and z variables, and D is a constant term.

2. How do you find the equation of a plane?

To find the equation of a plane, you need to know the coordinates of at least three points that lie on the plane. Then, you can use these points to set up a system of equations and solve for the coefficients A, B, C, and D. Another method is to use the normal vector of the plane and a point on the plane to create the equation.

3. What information do you need to find the equation of a plane?

To find the equation of a plane, you need to know the coordinates of at least three points that lie on the plane or the normal vector of the plane and a point on the plane. It is also helpful to know the general form of the equation, which is Ax + By + Cz + D = 0.

4. Can you find the equation of a plane if you only know two points?

No, you need at least three points to find the equation of a plane. This is because two points only determine a line, not a plane. However, if you know the normal vector of the plane and a point on the plane, you can find the equation with only two pieces of information.

5. What is the normal vector of a plane?

The normal vector of a plane is a vector that is perpendicular to the plane and points in the direction of the plane's surface. It is typically denoted as a unit vector with components (A, B, C), where A, B, and C are the coefficients of the equation Ax + By + Cz + D = 0. The normal vector is important in finding the equation of a plane because it can be used to determine the coefficients of the equation.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
8
Views
1K
  • Precalculus Mathematics Homework Help
Replies
17
Views
908
  • Precalculus Mathematics Homework Help
Replies
7
Views
521
  • Precalculus Mathematics Homework Help
Replies
2
Views
1K
  • Precalculus Mathematics Homework Help
Replies
7
Views
824
  • Precalculus Mathematics Homework Help
Replies
9
Views
2K
  • Precalculus Mathematics Homework Help
Replies
3
Views
2K
  • Precalculus Mathematics Homework Help
Replies
3
Views
2K
  • Precalculus Mathematics Homework Help
Replies
2
Views
1K
  • Precalculus Mathematics Homework Help
Replies
2
Views
1K
Back
Top