Equation of a plane given point and line in parametric form

In summary: So it is not a valid equation of the plane.In summary, the problem is asking to find the equation of a plane that contains the point (-1,0,1) and the vector <3t,t,8>. However, this information is not enough to uniquely determine a plane. The attempted solution involved finding a normal vector using the cross product, but the resulting equation of the plane (2x+2x=-1) is not valid as it does not contain the given point. Further information is needed to accurately solve this problem.
  • #1
musicmar
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0

Homework Statement


Find the equation of the plane that contains P=(-1,0,1) and r(t)=<3t,t,8>



Homework Equations





The Attempt at a Solution



n * <r-r0>=0
n * <t+2, 2t, 3t> = 0

I distributed the n, adding the terms and obtained:

1/t = -2n(n+2n+3n)

Clearly, I've done something wrong. If someone could point me in the right direction with even how to start this problem correctly, that would be great.

Thanks!
 
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  • #2
Ok, I set t=1, found a point P0 (2,2,2) and found two vectors OP and OP0 and took the cross product.
I then took the dot product of the normal vector I found above and the given point P.

So, my equation of the plane is 2x + 2x = -1

Can someone tell me if this is valid/and/or correct?
 
  • #3
musicmar said:

Homework Statement


Find the equation of the plane that contains P=(-1,0,1) and r(t)=<3t,t,8>
You won't be able to get a unique plane if all you know is a point in the plane and a vector that lies in it.
musicmar said:

Homework Equations





The Attempt at a Solution



n * <r-r0>=0
n * <t+2, 2t, 3t> = 0
I'm not following what you're doing above. I get that n is a normal to the plane, but where did <t + 2, 2t, 3t> come from?
musicmar said:
I distributed the n, adding the terms and obtained:

1/t = -2n(n+2n+3n)
?
musicmar said:
Clearly, I've done something wrong. If someone could point me in the right direction with even how to start this problem correctly, that would be great.

Thanks!
 
  • #4
2x+ 2x= -1?? The plane x= -1/4 is the plane parallel to the yz-plane at x= -1/4.
Did you mean 2x+ 2y= -1 or perhaps 2x+ 2z= -1?

musicmar, a line and a point not on the line determine a plane. But a "vector" is not a "line". It gives the direction but not specific points. Just given a "vector" we might have a line in the direction of the vector through the given point- and that will not determine a plane.

Anyway, to check if 2x + 2x= 4x = -1 is a solution, see if it meets the conditions. Does the given point (-1, 0, 1) lie in it? No, it doesn't; [itex]4(-1)\ne -1[/itex]. Nor does it lie in 2x+ 2y= -1 or 2x+ 2z= -1. [itex]2(-1)+ 2(0)= -2\ne -1[/itex] and [itex]2(-1)+ 2(1)= 0\ne -1[/itex].
 

1. What is an equation of a plane given a point and a line in parametric form?

The equation of a plane can be written in the form Ax + By + Cz + D = 0, where A, B, and C are the coefficients of the variables x, y, and z, and D is a constant term. To find this equation, we need to use the coordinates of the given point and the direction vector of the given line.

2. How do you find the direction vector for a line in parametric form?

The direction vector for a line in parametric form is given by the coefficients of the parameters t, u, and v. For example, if the line is represented by the equations x = 2t + 1, y = 3u + 2, z = 4v + 5, then the direction vector is v = (2, 3, 4).

3. Can you explain the process of finding the equation of a plane using a point and a line in parametric form?

First, we need to find the direction vector of the given line. Then, using the coordinates of the given point and the direction vector, we can construct two vectors that lie on the plane. These vectors will be perpendicular to the normal vector of the plane. Finally, we can use the cross product of these two vectors to find the normal vector and then use it to write the equation of the plane.

4. How many solutions are there for the equation of a plane given a point and a line in parametric form?

There is only one unique solution for the equation of a plane given a point and a line in parametric form. This is because a point and a line determine a unique plane in 3-dimensional space.

5. Can the equation of a plane be written in other forms besides the standard form Ax + By + Cz + D = 0?

Yes, the equation of a plane can also be written in vector form as N · (P - P0) = 0, where N is the normal vector, P is any point on the plane, and P0 is a known point on the plane. It can also be written in parametric form as P = P0 + sV1 + tV2, where P is any point on the plane, P0 is a known point on the plane, and V1 and V2 are two vectors lying on the plane.

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