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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.
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)
or maybe more simple: find the equation of a line through 2 of the points, and show the 3rd point does not satisfy the equation. So you know all 3 points do not lie on the same line, and therefore define a plane.
What's an equation of a line?
I'm just looking thru my textbook here, but I can only see equation of a plane. I'll use Google now though.
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?
B-A points in the direction from A to B |dw:1333072072947:dw| you scale it to move along it. Add A so the direction vector starts at A rather than the origin so A+ n(B-A)
*should be B-A as the label
Ok. Maybe I should ask what I should actually do for this question then. I don't seem to be able to comprehend our conversation.
I just solve for that? All I had to do was plug them in at random onto a vector equation? Sorry, my eyes are starting to hurt.
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.
Here is the equation of the line through points (2,1,5) and (-1,-1,10) (2,1,5) + n( -1 -2, -1-1, 10-5) (2,1,5) +n( -3, -2, 5) Is there an n that gets us to the point (8,5,-5)? n= -2 gives us (2,1,5)-2(-3,-2,5) = (2+6, 1+4,5-10)= (8,5,-5) so it looks like there are all on the same line.
( -1 -2, -1-1, 10-5) I'm confused about that. I was never taught to do that in class. Also, "Is there an n that gets us to the point (8,5,-5)? n= -2 gives us (2,1,5)-2(-3,-2,5) = (2+6, 1+4,5-10)= (8,5,-5)" Do I figure out n's value by guess and check?
No you don't guess. you have 3 separate equations for x,y,z example 2+n(-3) = 8 is the first one. It requires n= -2. it turns out all of the equations work for -2. so point (8,5,-5) is on the line
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?
subtracting the 2 points. Maybe 2-d is easier? say you are at (1,1) and you want to get to (2,2) how much to you move in the x and y? subtract to find you have to move (1,1) the (1,1) represents your slope. to get from one point on the line to another move over 1 and up 1. Or scale the 1,1 to get to any point on the line. example you are at 1,1 and you move 1/2 over and 1/2 up. You are still on the line
Ok. That makes sense I guess. I hope.
To continue the 2-d example. if we say (x,y)= n(1,1) this means all points where y=x Say we want the line y= x+1 then we could say (x,y)= (0,1)+n(1,1)
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.