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.(adsbygoogle = window.adsbygoogle || []).push({});

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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# 3 Points in a line!

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