I have a Triangle ABC:
$$A=\pmatrix{1\\0\\2} B= \pmatrix{3\\8\\0} C= \pmatrix{0\\7\\5}$$
Now my task is to get the median line that splits AB. I thought in getting it like this:
$$s = C + 0.5 * AB$$
So i get:
$$\pmatrix{1\\11\\3}= \pmatrix{0\\7\\5}+ 0.5*\pmatrix{2\\8\\-2}$$
But thats wrong the solution should be:
$$s = AC - 0.5 *AB$$
$$\pmatrix{-2\\3\\4}= \pmatrix{-1\\7\\3}+ 0.5*\pmatrix{2\\8\\-2}$$
Can somebody explain to me why i have to use AC as Vector for rhe middle line? Thanks
I interpret your notation $AC$ to mean a vector parallel to the line through $A$ and $C,$ where the vector magnitude is the distance from $A$ to $C$ and the vector direction is the direction from $A$ to $C.$ You could say this is "the vector from $A$ to $C,$" though this obscures the fact that there are many other pairs of points you could choose such that the same vector goes "from" the first point "to" the second point.
It makes things a little confusing when we use the same notation for a vector that we do for a point. But using this notation, you can compute a vector by taking the sum or difference of two vectors, a multiple of a vector, or the difference of two points. A point can be obtained by adding a vector to another point.
It appears that you're looking for the vector $MC$ where $M$ is the midpoint of the segment from $A$ to $B.$
One way to compute $MC$ is to first find the midpoint $M:$ $$ M = A + \frac12 AB = \pmatrix{2\\4\\1}.$$ Then $$ MC = C - M = \pmatrix{-2\\3\\4}.$$ Alternatively, using the facts that $AC = C - A$ and $AC = AB + BC,$ $$ A = C - AC, $$ $$ AB = AC - BC, $$ $$ M = (C - AC) + \frac12(AC - BC) = C - \frac12(AC + BC), $$ and so $$ MC = C - M = \frac12(AC + BC). $$ This is nice and symmetric. It does involve the vector $AC$ but also the vector $BC,$ which is appropriate since you're trying to "split the difference" between the two sides of the triangle.
Because you can write $MC = \frac12(AC + BC),$ and because $BC = AC - AB,$ you can also write $$MC = \frac12(AC + (AC - AB)) = \frac12(2 AC - AB) = AC - \frac12 AB. $$ But for similar reasons you could just as well write $$MC = BC + \frac12 AB. $$ There's nothing about the vector $AC$ that makes it better than the other vectors.