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I am working on a vision application and therefore I need to find the $x$ and $y$ values when $z=0$ for an object that is moving linearly (constant speed).

So what would $x$ and $y$ for point 3 be, if we know that $z=0$ and the speed between $2\to 3$ is equal to the speed between $1\to2$.

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    The speed of the object is irrelevant here. All that matters is that it’s moving in a straight line. – 2017-01-15

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Set up the movement equation: $$\begin{pmatrix}x(t)\\y(t)\\z(t)\end{pmatrix}=\vec{x}(t)=\vec{x_0}+t\cdot \vec{d}=\begin{pmatrix}x_0\\y_0\\z_0\end{pmatrix}+t\cdot \begin{pmatrix}d_x\\d_y\\d_z\end{pmatrix},\tag{1}$$ where $\vec{d}$ is the direction of the object and $\vec{x_0}$ any point it travels through.

Now solve $z(t)=0$ for $t$, i.e. solve $$z(t)=z_0+t\cdot d_z=0$$ and plug in back on the right hand side of (1) to get the values of $x$ and $y$.

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    Is this true? d = point 2 - point 1 E.g. p1(1,2,3) p2(3,5,7) d = 3-1,5-2,7-3 d = (2,3,4) – 2017-01-15
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    yes. Thats indeed true. – 2017-01-15
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    So when I take (1,2,3) as (x0, y0, z0), 0.4 as t and (2,3,4) as (dx, dy, dz) it gives me a vector (1,8;2,2;4,6). Is this correct? If so, what's the next step? – 2017-01-15
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    The first example you have done is correct. Now you need to let $t$ vary, and find what value it must take so that the $z$ component is zero... – 2017-01-15
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    I think there is a small error in your movement equation. Please take a look at (x0, y0,y0) on the right side. – 2017-01-15
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    I think I got strange answers because my points where increasing instead of decreasing. Could you help me solving this new example? Assume I have 2 points p0(5, 8, 11) and p1(3, 5, 7), d would be (-2, -3, -4). At t=0.4, my vector becomes (4.2, 6.8, 9.4). 11 + t * -4 --> t = 11/4. When I plug in 11/4 in movement equation, I get (-1/2, -1/4, 0) as an answer. – 2017-01-15
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    Your answer is correct :) – 2017-01-15
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    So vector (4.2, 6.8, 9.4) isn't part of the calculations? Just a vector at t=0.4?I thought I needed it to calculate something from it :) – 2017-01-15
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    exacly. Nono, there is nothing to be done with that vector – 2017-01-15