Why does the straight line from $(x_1,+y_1,+z_1)$ to $(x_2,+y_2,+z_2)$ become $r(\vec t)=(1-t)(x_1,+y_1,+z_1)+t(x_2,+y_2,+z_2)$ for $0 \leq t \leq 1$?
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You can write it this way: this is the line that goes though the point $A_1(x_1,y_1,z_1)$ and that follows the vector $A_1A_2$ of coordinates $(x_2-x_1,y_2-y_1,z_2-z_1)$.
Thus
$x=x_1+t(x_2-x_1)$, $y=y_1+t(y_2-y_1)$, $z=z_1+t(z_2-z_1)$, $t\in \mathbb{R}$
EDIT: if you want to be strictly between $A_1$ and $A_2$ (EDIT2: not going beyond, but the two points being still included), then $t\in [0,1]$
Martigan
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May be the interval should be open? – Bernard Jun 01 '15 at 08:27
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I ill used the term strictly. I meaned to include both $A_1$ and $A_2$. Otherwise yes, it should be open. – Martigan Jun 01 '15 at 08:29