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3 Points in a line!

by sutupidmath
Tags: line, points
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sutupidmath
#1
Jan14-09, 10:47 PM
P: 1,633
I encountered a problem where it asked us to show whether 3 points belong to a straight line in 3D. I have worked such problems in 2D before, but never in 3D, so i tried to carry out the same idea here as well. I understand that to do such a thing as a way one could start by first deriving the eq. of such a line that connects two points,(using some vector algebra i would think), and then test whether the coordinates of the third point satisfy the equations, however i am interested to know whether the following method would work as well. In fact it worked for 2 or 3 cases that i tested, but i don't know how would i show that it is valid in general.


I started by assuming that since we are working in 3D, we are gonna have slopes with respect to the x-axis and y-axis as well. Say we have three points in space:

[tex]P_1(x_1,y_1,z_1),P_2(x_2,y_2,z_2),P_3(x_3,y_3,z_3)[/tex]

Now the slopes with respect to x-axis and y-axis respectively would be:

(i) for the segment(line)

[tex]\bar{P_1P_2}[/tex]

[tex]\frac{z_2-z_1}{x_2-x_1}, \frac{z_2-z_1}{y_2-y_1}[/tex]

(ii) for [tex] \bar {P_1P_3}[/tex]


[tex]\frac{z_3-z_1}{x_3-x_1},\frac{z_3-z_1}{y_3-y_1}[/tex]


(iii) for [tex]\bar {P_2P_3}[/tex]


[tex]\frac{z_3-z_2}{x_3-x_2},\frac{z_3-z_2}{y_3-y_2}[/tex]


Now, in order for these points to lie on a straight line, their slopes w.r.t x and y have to be equal.

Also, after we set them equal, i managed with some elementary algebraic manipulations to come with a single relation, which worked for those same cases i tested:


[tex]\frac{z_2-z_1}{z_3-z_1}=\frac{y_2-y_1}{y_3-y_1}=\frac{x_2-x_1}{x_3-x_1}[/tex]


THis last expression, indeed, turned out to be not the only one, with some slightly different manipulatoins i could come up with simmilar relations which seemed to work as well.


So any feedback?

THnx in advance.
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soandos
#2
Jan14-09, 10:51 PM
P: 166
try this:
pick any two points
find the equation
see if the third point fits.

i think that is easier than what you did.
vector algebra is not needed.
just take two points, (x_1,y_1,z_1),(x_2,y_2,z_2).
not really sure how this is supposed to work in 3D, but i would find 2 different slopes, one in the xy plane, and one in the yz plane, and just work it out that way.
HallsofIvy
#3
Jan15-09, 06:53 AM
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The line through the points [itex](x_1, y_1, z_1)[/itex] and [itex](x_2, y_2, z_2)[/itex] is given by
[itex]x= x_1+ (x_2-x_1)t[/itex]
[itex]y= y_1+ (y_2-y_1)t[/itex]
[itex]z= z_1+ (z_2-z_1)t[/itex]

If t= 0, that gives [itex](x_1, y_1, z_1)[/itex] and when t= 1, [itex](x_2, y_2, z_2)[/itex]. Since a line is determined by two points that is all you need.

Now, to determine whether [itex](x_3, y_3, z_3)[/itex] is on that line, try to solve:
[itex]x_3= x_1+ (x_2-x_1)t[/itex]
[itex]y_3= y_1+ (y_2-y_1)t[/itex]
[itex]z_3= z_1+ (z_2-z_1)t[/itex]
See if that can be solved for a single value of t. Solve, say, the first equation for t and the see if it also satisfies the other two equations. If so the point is on the line, if not, the point is not on the line.

dodo
#4
Jan15-09, 10:40 AM
P: 688
3 Points in a line!

Perhaps another possibility is to check if the cross product of the vectors P1P2 and P1P3 is the zero vector. But Halls' method looks neater.
HallsofIvy
#5
Jan15-09, 11:15 AM
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Actually, Dodo, I think your suggest is simplest.


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