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Given $P_1=(1, 1, 1)$, $P_2=(2, -1, 2)$ and $P_3=(3,0,1)$, I need to prove that these three points are not on the same line.

What I tried - I showed that $\vec{P_1P_2}$, $\vec{P_1P_3}$ and $\vec{P_2P_3}$ are different (meaning they have different directions?). Does that conclude the proof?

Thanks.

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The points $P_1$, $P_2$ and $P_3$ are not on the same line iff $\vec{P_1P_2}$ and $\vec{P_1P_3}$ are not colinear (ie. there exists $\lambda \in \mathbb{R}$ such that $\vec{P_1P_2} = \lambda \vec{P_1P_3}$).

For example, $A=(0,0,0)$, $B=(1,1,1)$ and $C=(2,2,2)$ are on the same line, but $\vec{AB}$ and $\vec{AC}$ are different; however, $\vec{AC}=2 \vec{AB}$.

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  • $\begingroup$ Language difficulties... could you explain what colinear means? $\endgroup$ – homiee Apr 2 '13 at 8:20
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    $\begingroup$ @homiee: I added the definition. $\endgroup$ – Seirios Apr 2 '13 at 8:23
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Let slope of $\vec{P_1P_2} = m_1$ and slope of $\vec{P_2P_3} = m_2$, you see that $m_1 \neq m_2$.

Aliter:

$|P_1P_3|=|P_1P_2|+|P_2P_3| \implies $ The points are in straight line,else they are not.

Here $|xy|$ denotes the distance from point $x$ and $y$.

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