# Determining a path to represent a plane?

• Karnage1993
I'll try to figure out how to use the Heaviside function for the function in my original post.In summary, the goal is to find a path that represents a triangle with vertices (1, 0, 0), (0, 2, 0), and (0, 0, 3) by using parametric equations. One way to do this is to split the interval from 0 to 3 into three parts, with each part representing a different side of the triangle. Another method is to use the Heaviside function to create a piecewise function that combines the three individual parametric equations. Ultimately, there are multiple ways to parametrize the path and it is up to the individual to choose the
Karnage1993

## Homework Statement

Find a path $\gamma (t)$ representing the triangle (1, 0, 0) to (0, 2, 0) to (0, 0, 3) to (1, 0, 0)

None

## The Attempt at a Solution

Well I followed the steps on how to get an equation of a plane that contains the 3 vertices and I got $6x + 3y + 2z - 6 = 0$.

Solving for z, I have $z = 3 - (3/2)y - 3x$ so the path along the triangle can be represented as:

$f(x,y) = (x, y, 3 - 3x - (3/2)y)$

But this is not a function of t. How do I make it only a function of t? Do I set both x and y = t so that $\gamma (t) = (t, t, 3 - 3t - (3/2)t)$?

Karnage1993 said:

## Homework Statement

Find a path $\gamma (t)$ representing the triangle (1, 0, 0) to (0, 2, 0) to (0, 0, 3) to (1, 0, 0)

None

## The Attempt at a Solution

Well I followed the steps on how to get an equation of a plane that contains the 3 vertices and I got $6x + 3y + 2z - 6 = 0$.

Solving for z, I have $z = 3 - (3/2)y - 3x$ so the path along the triangle can be represented as:

$f(x,y) = (x, y, 3 - 3x - (3/2)y)$

But this is not a function of t. How do I make it only a function of t? Do I set both x and y = t so that $\gamma (t) = (t, t, 3 - 3t - (3/2)t)$?

What you have described is a parametric equation of the plane containing the three points. Surfaces need two parameters which, in your example, are x and y. You don't need the plane to parameterize the boundary. Remember if P and Q are two points, then the vector equation of the line from P to Q is$$\vec \gamma(t) = (1-t)P + tQ$$
The point will move from P to Q as t goes from 0 to 1. You will need three pieces to represent the three sides.

Hmm, I don't think we make line equations like that in my class. We do it in the form l(t) = p + tv, where p is one of the two points and v is the direction vector.

Finding the direction vectors for each, I get:

Let $$\vec v_1 = (0,2,0) - (1,0,0) = (-1,2,0)$$
$$\vec\gamma_1(t) = (1,0,0) + t(-1,2,0)$$

Let $$\vec v_2 = (0,0,3) - (0,2,0) = (0,-2,3)$$
$$\vec\gamma_2(t) = (0,2,0) + t(0,-2,3)$$

Let $$\vec v_3 = (1,0,0) - (0,0,3) = (1,0,-3)$$
$$\vec\gamma_3(t) = (0,0,3) + t(1,0,-3)$$

Now that I have line equations for each part of the triangle, what would I do to combine them to make ##\vec\gamma(t)##?

Karnage1993 said:
Hmm, I don't think we make line equations like that in my class. We do it in the form l(t) = p + tv, where p is one of the two points and v is the direction vector.

Finding the direction vectors for each, I get:

Let $$\vec v_1 = (0,2,0) - (1,0,0) = (-1,2,0)$$
$$\vec\gamma_1(t) = (1,0,0) + t(-1,2,0)$$
Okay, this goes from (0, 2, 0) to (1, 0, 0) as t goes from 0 to 1. If you were to replace t with 3s, getting $\vec{\gamma(s)}= (1, 0, 0)+ 3s(-1, 2, 0= (1, 0, 0)+ s(-3, 6, 0)$, you have a vector function that goes from (0, 2, 0) to (1, 0, 0) as s goes from 0 to 1/3.

Let $$\vec v_2 = (0,0,3) - (0,2,0) = (0,-2,3)$$
$$\vec\gamma_2(t) = (0,2,0) + t(0,-2,3)$$
Okay, this goes from (0, 2, 0) to (0, 0, 3) as t goes from 0 to 1. Since we already have "from (0, 2, 0) to (1, 0, 0)", in order to be "moving" around the triangle, we would prefer to go from (0, 0, 3) to (0, 2, 0). We could do that by replacing t with r= 1- t, with r from 0 to 2. But, further, because that will attach to the previous segment, we want it to end the chain- that is we want to have the parameter from 2/3 to 1. s will go from 2/3 to 1 as r goes from 0 to 2 if s= 2/3+ (1/3)r which is the same as r= 3(s- 2/3)= 3s- 2 . That means that we must have t= 1- (3s- 2)= 2- 3x. $\vec{\gamma(s)}= (0, 2, 0)+ (2- 3s)(0, -2, 3)= (0, 0, 6)+ (0, 6s, 9s)$ for s from 2/3 to 1.

Let $$\vec v_3 = (1,0,0) - (0,0,3) = (1,0,-3)$$
$$\vec\gamma_3(t) = (0,0,3) + t(1,0,-3)$$
Okay, this goes from (1, 0, 0) to (0, 0, 3) as t goes from 0 to 1 and this side is the middle of the chain. We want it to go from (1, 0, 0) to (0, 0, 3) as s goes from 1/3 to 2/3 so we want t= 3(s- 1/3)= 3s- 1. $\vec{\gamma(s)}= (0, 0, 3)+ (3s-1)(1, 0, -3)= (-1, 0, 6)+ (3s, 0, -9s)$ for s from 1/3 to 2/3.

Now that I have line equations for each part of the triangle, what would I do to combine them to make ##\vec\gamma(t)##?
I you don't like piecewise defined functions, you can use the "Heaviside function", H(x) which is 0 for x< 0, 1 for $x\ge 0$. If F(x)= f(x) for x< a, g(x) for $a\le x< b$, and h(x) for $b\le x$, then you can write
F(x)= f(x)+ H(x- a)(g(x)- f(x))+ H(x- b)(h(x)- g(x)).

Last edited by a moderator:
EDIT: I think I've figured out why you split the interval into three.

So the piecewise function would look like:

$$\vec\gamma(s) = \begin{cases} (1,0,0)+s(−3,6,0), & 0 \le t < \frac{1}{3}\\ (−1,0,6)+s(3,0,−9), & \frac{1}{3} \le t < \frac{2}{3}\\ (0,0,6)+s(0,6,9), & \frac{2}{3} \le t \le 1 \end{cases}$$

Last edited:
Karnage1993 said:
This replacing of t with a multiple of another variable is confusing me. I started to make a piecewise function for the path but I ended up with:

$$\vec\gamma(t) = \begin{cases} (1,0,0) + t(-1,2,0), & 0 \le t \le 1\\ (0,2,0) + t(0,-2,3), & 0 \le t \le 1\\ (0,0,3) + t(1,0,-3), & 0 \le t \le 1 \end{cases}$$

which is a problem because I want to have the second line "activate" as soon as the first line is done tracing it's path. Should I use some sort of switch function that moves onto the next line vector equation (ie, as soon as I reach the point (0,2,0))?

In your second equation replace ##t## by ##t-1## and let ##t## go from 1 to 2. Similarly in the third equation so the result is ##t## varying from 0 to 3.

Thank you both. I see now that there are infinitely many ways to parametrize the path. I think the most easiest would be to go from 0 to 1, 1 to 2, then 2 to 3 as LCKurtz described it.

## 1. What is a plane?

A plane is a flat surface that extends infinitely in all directions. In mathematics, a plane is represented by a two-dimensional coordinate system, with two axes intersecting at right angles.

## 2. How do you determine a path to represent a plane?

To determine a path to represent a plane, you can use the following steps:

1. Identify two distinct points on the plane.
2. Find the equation of the line passing through these two points.
3. Repeat this process for another set of two points on the plane.
4. The intersection of these two lines will give the third point on the plane.
5. Use the three points to plot a triangle, which will represent the plane.

## 3. Can a plane be represented by a single point?

No, a plane cannot be represented by a single point. A plane is a two-dimensional object and requires at least three points to be defined.

## 4. What is the purpose of determining a path to represent a plane?

Determining a path to represent a plane is important for many mathematical and scientific applications. It allows us to visualize and analyze the properties of a plane, such as its slope, intercepts, and distance from other objects.

## 5. Are there different ways to represent a plane?

Yes, there are multiple ways to represent a plane. In addition to the path method described above, a plane can also be represented by its normal vector and a point on the plane, or by its equation in the form of Ax + By + Cz + D = 0 where A, B, and C are the coefficients of the plane's normal vector and D is a constant term.

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