$1a)$ $a$ and $b$ are position vectors of points A and B on the plane. Show that the line that passes through A and B has the vector equation $r= sa - (s-1)b$
$1b)$ What value of s does the point P with position vector $p$ lie between points A and B on the line.
I know that the equation of a plane is shown by $r=r0 + sa + tb$ , but do not understand what the question is asking.
A vector beginning at $B$ and ending at $A$ is given by ${\bf a}-{\bf b}$ (end minus start). Thus for any scalar $s$, $({\bf a}-{\bf b})s$ is parallel to the vector ${\bf a}-{\bf b}$ (and thus is parallel to the line through $A$ and $B$. Now to get the line through $A$ and $B$ you merely need to shift ${\bf r}(s) = ({\bf a}-{\bf b})s$ away from the origin. Since it needs to pass through $B$, let's add ${\bf b}$. This gives us $${\bf r}(s) = {\bf b}+({\bf a}-{\bf b})s = {\bf a}s-(s-1){\bf b}$$
Alternatively, just notice that when $s=0$ we get ${\bf r}(0)=0{\bf a}-(0-1){\bf b}={\bf b}$ and when $s=1$ we get ${\bf r}(1)=1{\bf a}-(1-1){\bf b}={\bf a}$. So the line passes through $A$ (when $s=1$) and $B$ (when $s=0$).
So answer to the second question is $0 \leq s \leq 1$ (starting at $B$ running over to $A$).