## Linear Algebra: A Modern Introduction

$$\overrightarrow{AB} = b - a$$ $$\overrightarrow{BC} = - a$$ $$\overrightarrow{AD} = - 2a$$ $$\overrightarrow{CF} = 2a - 2b$$ $$\overrightarrow{AC} = b - 2a$$ $$\overrightarrow{BC} + \overrightarrow{DE} + \overrightarrow{FA} = \mathbf{0}$$
From the triangle $OAB$, we see that $\overrightarrow{OA} + \overrightarrow{AB} + \overrightarrow{BO} = 0$, so $a + \overrightarrow{AB} - b = 0$. Rearranging, we get $\overrightarrow{AB} = b - a$. Since $OABC$ forms a parallelogram, $\mid \overrightarrow{OA} \mid = \mid \overrightarrow{BC}\mid$, but the vectors are in opposite directions, so $\overrightarrow{BC} = - a$ Going from $A$ to $D$ requires twice the distance of traveling from the origin to $A$, but in the opposite direction. Thus, $\overrightarrow{AD} = - 2a$. Since $OBC$ forms a triangle, it follows that $\overrightarrow{CO} = -\overrightarrow{OB}-\overrightarrow{BC} = -b - (-a) = a -b$. Since $\overrightarrow{CF} = 2\overrightarrow{CO}$, $\overrightarrow{CF} = 2a - 2b$. Since $\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC}$ by the Head-to-Tail Rule, $\overrightarrow{AC} = (b-a) + (-a) = b - 2a$ Let $z = \overrightarrow{BC} + \overrightarrow{DE} + \overrightarrow{FA}$. Because $\overrightarrow{BC}$ has no $y$ component (it's horizontal), and the $y$ components of $\overrightarrow{DE}$ and $\overrightarrow{FA}$ are exactly opposite, the $y$ component of $z$ is $0$. The $x$ component of $\overrightarrow{BC}$ is equal and opposite the $x$ component of $\overrightarrow{DE} + \overrightarrow{FA}$, so the $x$ component of $z$ is $0$. Thus, $\overrightarrow{BC} + \overrightarrow{DE} + \overrightarrow{FA} = \mathbf{0}$.