I’m trying to understand vectors, and I have got part 1, but how would you express EF, is that also just 6b-6a?
[Math] Simple vectors question with hexagon
vectors
Related Solutions
The condition asks that any pairs out of your choice of vectors should have angle between them greater than $\pi/2$. This gives us certain geometric restraints on how vectors should be organized.
Let $v_1 \in \Bbb{R}^d \setminus \{0\}$ be our first choice and $H_1 = \{ u \in \Bbb{R}^d : u \cdot v_2 \geq 0 \}$. Then our next choice $v_2$ cannot be drawn from this region. But $H_1$ is a closed half-space, which means that half of the $\Bbb{R}^d$ becomes a forbidden region. And as you keep picking vectors, this forbidden region keep expanding. The problem says that when $d = 2$, if you pick 3 vectors in this way, the forbidden region covers the entire plane $\Bbb{R}^2$ and thus you cannot pick the fourth one.
Here is another explanation which is specific to the 2D case. Assume that you have four vectors. Pick any of them and label it $v_1$, and then label the others by $v_2, v_3, v_4$ as they appear in counter-clockwise order beginning from $v_1$. Then the angles $\angle v_1v_2$, $\angle v_2v_3$, $\angle v_3v_4$ and $\angle v_4v_1$ should sum up to $2\pi$. This means that at least one of them should be $\leq \pi/2$.
At each point in the space corresponds a position vector and viceversa:
$$P(x,y)\equiv OP=x \vec i+y\vec j$$
When you consider two points
$$P(a,b)\equiv OP=a \vec i+b\vec j \quad Q(c,d)\equiv O Q=c \vec i+d\vec j$$
the vector from $P$ to $Q$ is given by
$$PQ=P-Q=(a-c,b-d)=(a-c) \vec i+(b-d)\vec j$$
From a mathematical point of view vector $PQ$ is also a vector form the origin (in a vector space all vectors are form the origin)
$$PQ=OR=(a-c,b-d)=(a-c)\vec i+(b-d)\vec j$$
such that $$OP+OR=OQ$$
Anyway you could also though to $PQ$ as a vector applied in $P$ such that:
$$OP+PQ=OQ$$
and in this sense they can be moved like free vectors.
Best Answer
Since it is a regular hexagon, the vector from $E$ to $F$ is the same as the vector from $O$ to $A$. So we have $\overrightarrow{EF}= 6a$.