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use elimination to find if the three points are independent (not collinear)

Elimination?
-I think no, because there is no position vector or a direction vector.

Pi?

Sorry if I sound persistent, but I still don't understand (I have looked through the link).

what kind of math are you studying?

Vectors, of Vectors and Calculus.

Specifically, Lines and Planes.

Do they teach you about independent vectors? How to tell if two vectors are independent?

I had difficulties understanding the course. I don't exactly know.

equations of lines is the lesson (specifically)

What's an equation of a line?

y=mx+b? How does that work for a 3-space point?

I think you do A + n(B-A)
e.g. (2,1,5)+n ((-1,-1,10)-(2,1,5))
where n is any value (a scalar)

You mean like a vector equation?
[x,y,z]=[xo,y0,zo]+t[a1,a2,a3]+s[b1,b2,b3]

But that wouldn't make sense to me...Aren't the ones with scalar multipliers direction vectors?

*should be B-A as the label

It looks like the 3 points are not collinear, so they define a plane

Wait, how do I determine collinear? And, do I just do as what I suggested 2 comments above?

See if this helps
http://tutorial.math.lamar.edu/Classes/CalcIII/EqnsOfLines.aspx

How far should I read into this?

Start at the paragraph below the ellipse.

Ok. I just figured out that the equation of the line is the same as the vector equation.

that is point (-1,-1,10) - (2,1,5) rewritten as ( -1 -2, -1-1, 10-5)

Look at Example 1 in Paul's notes

I don't even understand that...

IS he adding the 2 points or subtracting them?

Ok. That makes sense I guess. I hope.

Good luck. At least you have the answer, It is not a plane.

Ok. Thanks for the help, and putting up with my lack of knowledge on the topic.