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I am given the points $(1,2,3), (2,4,1)$ and $(x,y,2)$ and I am being asked to find some values for $x$ and $y$ such that these 3 points will be collinear in $\mathbb{R}^3$

Most of this topic is covered with videos and tutorials in $\mathbb{R}^2$ and my teacher did not cover this topic at all, and it is not discussed in the book.

So far I have found out that the points are said to be collinear when the two shorter distances summed equals the longer distance between two points.

I started with the equation $\sqrt{(x-1)^2 + (y-2)^2 + 1} + \sqrt{(5-x)^2 + (3-y)^2 + 1} = \sqrt{21}$

But with two variables and one equation this does not appear to be the correct approach. I also feel it is incorrect as the $\sqrt{21}$ comes from the distance between the two known points, but not necessarily the longest distance as either the first or second point may be the point in-between.

I tried writing a short program which is checking for points between the two known points, but it is not returning any results and I would rather not expand the program to find the solution as this technique will not be available on a test of course. This implies that one of the two known points is somewhere in the middle.

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If you know determinants, there is a characterization of collinearity by the vanishing of the dterminant of all coordinates. See here, next to the end, for the $\mathbb{R}^3$ case. – 1015 Apr 10 '13 at 4:03
up vote 4 down vote accepted

A simpler approach is to remember the parametric equation for a line in $\Bbb R^3$. It is (some point)+t(direction vector). The line through $(1,2,3)$ and $(2,4,1)$ is then $(1,2,3)+t(2-1,4-2,1-3)=(1,2,3)+t(1,2,-2)$. Now you need to find $x$ and $y$ such that $(x,y,2)$ is of this form. The third coordinate will let you find $t \ldots$

Added: you are correct with $t=1/2$. You missed adding in $(1,2,3)$, so the point is $(1,2,3)+(1/2,1,-1)=(3/2,3,2)$ and you can read off $x$ and $y$. The idea is that you describe a line by adding a point on the line to a vector of any length in the direction the line points. As a simple example, the line through $(1,0,0)$ and parallel to the $y$ axis is all the points $(1,t,0)$. The above would express it as $(1,0,0)+t(0,1,0)$

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I think this is a good answer, but I am lacking so much basic knowledge that it is hard to understand exactly what I am doing. My instinct tells me based on what you say that $3 + t(-2) = 2$ so $t = 1/2$ and $(1/2, 1, -1)$ satisfies the equation? – Leonardo Apr 10 '13 at 4:18
Thanks, I did add the point and was checking it but with no clue what I was doing, but I can figure out why it works soon thanks to your explanation and example. Thank you! – Leonardo Apr 10 '13 at 4:27

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