What is the Cartesian equation of the plane containing a given line and point?

In summary, the question is asking for the Cartesian equation of a plane containing the line x=3t , y =1+t , z=2-t and passing through the point (-1,2,1). The solution involves finding the normal vector to the plane and setting up equations using the given point and a point on the line. Another method is using the cross product of two vectors to find the normal and then setting up an equation with the given point. However, neither method has resulted in the correct answer.
  • #1
craka
20
0

Homework Statement


Question is
"The Cartesian equation of the plane containing the line x=3t , y =1+t , z=2-t and passing through the point (-1,2,1) is?"


Homework Equations



[tex]
\begin{array}{l}
n \bullet (r - r_0 ) = 0 \\
< n_1 ,n_2 ,n_3 > \bullet < x - x_0 ,y - y_0 ,z - z_0 > = 0 \\
\end{array}
[/tex]


The Attempt at a Solution



direction vector is < 3 , 1, -1>

[tex]
\begin{array}{l}
< - 1,2,1 > \bullet < x - 3,y - 1,z + 1 > = 0 \\
- x + 2y + z = 2 \\
\end{array}
[/tex]

But this doesn't appear to be right. Could someone help me out here please. I'm at a lost on how to do this. Thanks
 
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  • #2


1. Pick a point ON the line, (f0,1,2) for example, and set up the equation governing this.
You have three unknowns (essentially, the components of the normal vector).
Utilize the fact that irrespective of the value of t, the whole line should be included in the plane.
This will give you a single equation for the three unknowns.
2. Also require that the given solitary point should lie in the plane.
This will give you the second equation for your three unknowns.

This is what you need, you'll end up with a free scaling parameter for the normal (i.e, its length), as you should.
 
  • #3


Or: Choose two points on the line, p0 and p1. Determine the vector from p0 to p1 and the vector from p0 to (-1, 2, 1). Take the cross product of those two vectors to find the normal to the plane.
 
  • #4


I still haven't been able to do this.

I tried to do with cross product of <0,1,2> x <-1,2,1>
to get vector normal , which was <-3,-2,1>

than did <-3,-2,1> . < x- (-1), y-2, z-1 >=0

which worked out to be -3x-2y+z=0 however this is not the answer still.
 

1) What is a Cartesian equation of a plane?

A Cartesian equation of a plane is a formula that represents the relationship between the x, y, and z coordinates of points on a plane. It is typically written in the form 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.

2) What is a line with parametric equations?

A line with parametric equations is a way of representing a line in three-dimensional space using parameters (usually denoted as t or s). The equations define the x, y, and z coordinates of points on the line in terms of these parameters. For example, the parametric equations for a line may be x = at, y = bt, and z = ct, where a, b, and c are constants.

3) How do you find the Cartesian equation of a plane containing a line with parametric equations?

To find the Cartesian equation of a plane containing a line with parametric equations, you can use the fact that any two points on the line can be used to define the plane. You can then use these points and the formula for a plane to find the coefficients A, B, C, and D. Alternatively, you can use the direction vector of the line and a point on the line to find the equation of the plane.

4) Can a point be used to determine the Cartesian equation of a plane containing a line?

Yes, a point can be used to determine the Cartesian equation of a plane containing a line. In fact, any point on the line can be used to define the plane. However, it is important to note that the point must be on the line and not just in the same plane as the line.

5) Are there any special cases when finding the Cartesian equation of a plane containing a line?

Yes, there are a few special cases when finding the Cartesian equation of a plane containing a line. One special case is when the line is parallel to one of the coordinate planes (x, y, or z). In this case, the Cartesian equation will have one of the coefficients (A, B, or C) equal to 0. Another special case is when the line is contained within the plane. In this case, the Cartesian equation will have all coefficients equal to 0, except for the constant D term.

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