# Find equation of plane given curve

• nicknack1016
In summary, someone was trying to find the normal vector to a plane, but they needed a point to find it. They found the normal vector when they found the derivative of the position vector. Then they used the directional vector to find the point where the tangent line hit the xy-plane.
nicknack1016

## Homework Statement

r(t) = (2t)i + (t^2)j + (1 - t^2)k
Find the equation of the plane that this curve lies on.

## The Attempt at a Solution

I've found planes when I had vectors or lines but now with a curve I'm not sure where to even start. Can someone just give me a starting point? I don't want it to be solved for me or anything. I'm just looking for a point to even begin.

Try eliminating the parameter from your equations

x = 2t, y = t2, z = 1 - t2.

The plane that r(t) lies on is always perpendicular to some vector, n. This vector, n, is normal to the plane. So find the vector, n, that is always normal to r(t). This vector defines the plane.

To find a normal vector wouldn't i need a point as well? Or can i just substitute in some value of T in r(T) to get a point?

gerben said:
The plane that r(t) lies on is always perpendicular to some vector, n. This vector, n, is normal to the plane. So find the vector, n, that is always normal to r(t). This vector defines the plane.

r(t) is a position vector from the origin to a point on the curve. The normal n to the plane will not be perpendicular to r(t).

solving for T in the equation x = 2t then substituting it into both of those equations gave me...
y = x^2/4, z = 1 - x^2/4

then using the Y equation
z = 1 - y then 1 = z + y
would this be my plane?

nicknack1016 said:
solving for T in the equation x = 2t then substituting it into both of those equations gave me...
y = x^2/4, z = 1 - x^2/4

then using the Y equation
z = 1 - y then 1 = z + y
would this be my plane?

Yes, it would. The fact that x is missing in the equation just means the plane is parallel to the x axis. x can be any value as long as y + z = 1. Now that you have the equation of the plane, it is a trivial matter to read off the normal.

Now I need to find the point where the tangent line at T = 2 intersects the xy-plane.
I found the derivative of r(t)..
r'(t) = 2i + (2t)j - (2t)k
then substituted in 2 for t
r'(2) = <2, 4, -4>
this is the directional vector of the tanget line.
how would i use this to find where it hits the xy-plane?

nicknack1016 said:
Now I need to find the point where the tangent line at T = 2 intersects the xy-plane.
I found the derivative of r(t)..
r'(t) = 2i + (2t)j - (2t)k
then substituted in 2 for t
r'(2) = <2, 4, -4>
this is the directional vector of the tanget line.
how would i use this to find where it hits the xy-plane?

You can figure out the position r(2) and you have the direction vector, so write the parametric equations of the line. Put the x,y, and z in the plane and figure out what parameter t works and...

wow really? haha awesome.
so r(2) = 4i + 4j - 3k
x = 4 + 2t, y = 4 + 4t, z = -3 - 4t
it intersects at z=0 sooo t = -3/4
when t = -3/4...x = 5/2, y = 1, and z = 0
=D thanks for all the help it was very much appreciated!

## 1. What is the equation of a plane?

The equation of a plane is a mathematical representation of a flat, two-dimensional surface in a 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.

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

To find the equation of a plane given a curve, you will need to know at least three points on the curve. These three points can be used to create a system of equations that can be solved to find the coefficients A, B, C, and D. Once you have these coefficients, you can write the equation of the plane in the form of Ax + By + Cz + D = 0.

## 3. What is the normal vector of a plane?

The normal vector of a plane is a vector that is perpendicular to the plane and has a magnitude of 1. It is often denoted by the symbol n and can be calculated using the cross product of two non-parallel vectors that lie on the plane. The normal vector is important in finding the equation of a plane, as it can be used to determine the coefficients A, B, and C.

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

No, you need at least three points on the curve to find the equation of a plane. This is because a plane is two-dimensional and can be rotated around an infinite number of axes, so two points are not enough to uniquely define it. With three points, you can create a system of equations and solve for the coefficients of the equation of the plane.

## 5. Is there a specific method for finding the equation of a plane given a curve?

Yes, there are several methods for finding the equation of a plane given a curve. One method is to use the normal vector of the plane and the coordinates of a point on the curve to create a system of equations and solve for the coefficients. Another method is to use vector calculus techniques, such as the gradient, to find the equation of the plane. The method used may vary depending on the complexity of the curve and the available information.

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