# Why doesn't this find the mid point?

I saw a simple question and decided to try an alternate method to see if I could get the same answer; however, it didn't work out how I had expected.

Given $A(4, 4, 2)~$ and $~B(6, 1, 0)$, find the coordinates of the midpoint $M$ of the line $AB$.

I realize that this is quite easy just taking $\frac{1}{2}(A+B) = (5, \frac{5}{2}, 1)$; however, I don't understand why this doesn't give me the same answer:

If I take $\frac{1}{2}\vec{AB}~$ I would have thought that I would be half way to B from A which would be the midpoint right? but, of course I get:

$\frac{1}{2}\vec{AB} = \frac{1}{2}(2, -3, -2) = (1, -\frac{3}{2}, -1)$

Is it just because this is a directional vector which doesn't indicate position in any way, and I am trying to halve the direction/angle or something?

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You computed half of the displacement from $A$ to $B$. Add this vector to $A$ to obtain the midpoint. – David Mitra Jun 3 '12 at 15:08

That's right...your calculation doesn't take into account position in any way. You are going half the distance from $A$ to $B$, but starting at the origin, not at $A$. Try $A+\frac{1}{2}\vec{AB}$
EDIT: It occured to me that I should point out: $$A+\frac{1}{2}\vec{AB}=A+\frac{1}{2}(B-A)=\frac{1}{2}(A+B)$$
The idea is a good one. But we need to add $\frac{1}{2}\overrightarrow{AB}$ to $\overrightarrow{A}$.
Remark: The idea can be generalized. Let $0 \lt t \lt 1$. We want to find the point $C$ on the line segment $AB$ such that the ratio $AC:DC$ is $t:1-t$. (You were dealing with the case $t=1/2$.) Then $$\overrightarrow{C}=\overrightarrow{A}+t\overrightarrow{AB}=\overrightarrow{A}+t(\overrightarrow{B}-\overrightarrow{A})=(1-t)\overrightarrow{A}+t\overrightarrow{B}.$$