Hello kullgirl418 Originally Posted by
kullgirl418 I am still confused as to how you are labeling B and C as vectors when they are representing points not lines?
It's clear that you're not really understanding vectors at all, so read this carefully.
I didn't label B and C as vectors - I referred to their position vectors: Originally Posted by
Grandad ... So if the position vectors of $\displaystyle B$ and $\displaystyle C$ are $\displaystyle \vec{b}$ and $\displaystyle \vec{c}$,...
...in other words, their positions in relation to some arbitrary origin - usually denoted by O.
If you're not clear what that phrase 'their positions in relation to some arbitrary origin' means, let me unpack it a little - but you'll have to do the work. So:
- Draw a triangle ABC - don't make it a special triangle; try not to make it right-angled or isosceles.
- Mark M as the mid-point of BC.
- Somewhere else on the paper mark a point O. It's easiest if it's outside the triangle and doesn't lie on any of its sides when they're produced.
Then the position vectors of the points A, B, C and M in relation to O are the displacement vectors $\displaystyle \vec{OA}, \vec{OB},\vec{OC}$ and $\displaystyle \vec{OM}$. (A displacement vector is a description of how you move from one point to another. So the displacement vector $\displaystyle \vec{OA}$ describes - using whatever language you'd like - how you would move from O to A.)
To make it simpler to write, we describe these position vectors - these 'movement descriptions' - using lower-case letters:$\displaystyle \vec{OA}=\vec{a}$
$\displaystyle \vec{OB}=\vec{b}$
$\displaystyle \vec{OC}=\vec{c}$
$\displaystyle \vec{OM}=\vec{m}$
The triangle law of addition is then simply two movement descriptions' combined into one. So when we write:$\displaystyle \vec{OA}+\vec{AB}=\vec{OB}$
we're just saying: $\displaystyle \underbrace{\text{a movement from O to A }}_{\vec{OA}}$ $\displaystyle \underbrace{\text{ followed by }}_+$ $\displaystyle \underbrace{\text{ a movement from A to B }}_{\vec{AB}}$ $\displaystyle \underbrace{\text{ is the same as }}_=$ $\displaystyle \underbrace{\text{ a movement from O to B}}_{\vec{OB}}$
So when we use lower-case letters this becomes:$\displaystyle \vec{a}+\vec{AB}=\vec{b}$
in other words$\displaystyle \vec{AB}=\vec{b}-\vec{a}$
which I described in an earlier reply to you as a very important result.
Finally, let's show you again where the 'mid-point' formula comes from.
Writing this important result again, but using the points B and M this time:$\displaystyle \vec{BM}=\vec{m}-\vec{b}$
But the movement from B to M is just one-half of the movement from B to C, since M is the mid-point of BC. So$\displaystyle \vec{BM}=\tfrac12\vec{BC}$
Knowing that $\displaystyle \vec{BC}=\vec{c}-\vec{b}$ (there's that result again), we can say:$\displaystyle \vec{BM}=\tfrac12\vec{BC}$
$\displaystyle \Rightarrow \vec{m}-\vec{b}=\tfrac12(\vec{c}-\vec{b})$
$\displaystyle \Rightarrow \vec{m}=\vec{b}+\tfrac12\vec{c}-\tfrac12\vec{b}$
$\displaystyle \Rightarrow \vec{m}=\tfrac12(\vec{b}+\vec{c})$
I hope that helps to make things clearer.
Grandad