Parametrization of a line formed by 3 points

In summary: However, in this case, the three points are co-linear and therefore all lie on the same line. This means that we can use any two of the points to parametrize the line.In summary, to parametrize the line formed by the points A, B, and P, you can choose any two points (e.g. A and P or B and P) and use the vector between them to create a parametrization. This parametrization will be in the form (x,y,z) = (x0,y0,z0) + k(vx,vy,vz) where k is an arbitrary scalar and (x0,y0,z0) is the starting point (e.g. A or B).
  • #1
forestmine
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Homework Statement



Find a parametrization of the equation of the line formed by the points A, B, and P.

A(2,-1,3) B(4,3,1) P(3,1,2)

Homework Equations



x=x_0+v_1*t
y=y_0+v_2*t
z=z_0+v_3*t

The Attempt at a Solution



Alright, so, I've already determined that P is equidistant from the points A and B, and therefore the center of a sphere. I'm having a hard time finding a parametrization since we're dealing with 3 points, as oppose to 2. If, for example, it asked for the parametrization of the equation of the line formed by A and P, I would find the vector AP, and then simply use A as my (x_0, y_0, z_0) and plug in the v-values as well. Since it's along the line formed by all three points, however, I'm a bit confused.

Finding the vector AP suggests a line parallel to the line I'm setting up parameters for, but what about a third point? The point BP would be parallel to said line as well, but can I assume it's along the same line as AP, or is the fact that it's parallel sufficient? Even so, how do I include B in this?

Hope this makes sense.

Any help would be greatly appreciated! Thank you!

EDIT// So I've just realized that PA and PB are anti-parallel. If that's the case, can I use just one or the other, and would either parametrization be correct?
 
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  • #2
Alright, so, I've already determined that P is equidistant from the points A and B, and therefore the center of a sphere.
Circle... with A and B on the circumference. But |AB| != |PA|? Of course all three could be on the circumference of a conic section?

Have you checked that A B and P are not co-linear?

What is the definition of a straight line in 3 dimensions? (i.e. why would you need three points - normally you'd just use two points that are on the line right?)
 
  • #3
A, B, and P are co-linear, right? And I'm saying that based on the fact that AP and BP are anti-parallel vectors.

And since they all lie along the same line, I wouldn't need three points, and so a parametrization using AP is just as valid as one using BP, even though they appear different?

I'm not completely comfortable with the concept of parametrizations, but it seems to me, these equations could describe the same line using any point along the line, and so it'd make sense that there would be various forms, right?

Hopefully I'm on the right track...thanks for the help! (:
 
  • #4
If A B and P are co-linear, then you just need two of them to construct a vector that points along the line. Any other point on the line will be some scale-factor along that vector from any other point on the line.

so construct a vector v from A to B (say) it will have components (vx, vy, vz) ... if A has components (u,v,w) then the line parametrizes to the set of points Q=(x,y,z) such that:

(x,y,z)=(u,v,w)+k(vx,vy,vz): k is an arbitrary scalar.

I'm not saying that A B and P are co-linear here - you have to check.
Usually three points are used to define a plane.
 

1. What is parametrization of a line formed by 3 points?

Parametrization of a line formed by 3 points is a mathematical process of defining a line in terms of one or more parameters. It involves finding an equation that represents the line passing through 3 given points.

2. What are the steps involved in parametrization of a line formed by 3 points?

The steps involved in parametrization of a line formed by 3 points are:

  1. Find the slope of the line using the formula (y2-y1)/(x2-x1).
  2. Choose any of the given points and plug in its coordinates along with the slope into the point-slope form of a line equation (y-y1)=m(x-x1).
  3. Substitute the values of the other two points into the equation to get two equations with two unknowns.
  4. Solve the equations to find the values of the parameters.
  5. Plug in the values of the parameters into the point-slope form equation to get the final parametric equation of the line.

3. Why is parametrization of a line formed by 3 points useful?

Parametrization of a line formed by 3 points is useful in various applications such as computer graphics, physics, and engineering. It allows for a more flexible representation of a line, making it easier to manipulate and calculate.

4. Can a line formed by 3 points have multiple parametric equations?

Yes, a line formed by 3 points can have multiple parametric equations. This is because there are infinite ways to parametrize a line passing through 3 points. However, all the parametric equations will represent the same line.

5. Is it possible to parametrize a line formed by 3 points in 3-dimensional space?

Yes, it is possible to parametrize a line formed by 3 points in 3-dimensional space. The process is similar to parametrization in 2-dimensional space, but it involves finding equations in terms of 3 parameters instead of 2.

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