Parametrization of a line formed by 3 points

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Homework Help Overview

The discussion revolves around finding a parametrization of the line formed by three points A, B, and P in three-dimensional space. The original poster expresses confusion regarding how to incorporate all three points into a single parametrization, given that typically only two points are needed to define a line.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the relationship between the points A, B, and P, questioning whether they are co-linear and discussing the implications of this on the parametrization. The original poster considers the vectors AP and BP and their parallelism, while others suggest that only two points are necessary if the points are indeed co-linear.

Discussion Status

The discussion is ongoing, with participants providing insights into the nature of the points and their relationships. Some guidance has been offered regarding the use of vectors to define the line, but there is no explicit consensus on the co-linearity of the points or the best approach to the parametrization.

Contextual Notes

There is uncertainty regarding the co-linearity of points A, B, and P, which is crucial for determining the correct parametrization. Participants also note the potential for different parametrizations based on the choice of points along the line.

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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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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?)
 
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! (:
 
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.
 

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