Chapter Vectors, Question Problem 14EQ

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, $\overrightarrow{BC}+\overrightarrow{DE}+\overrightarrow{FA}=0$.