Verify analytically and graphically the following vector…

Verify analytically and graphically the following vector equalities: $1^\circ\quad \vec{a}+\tfrac{1}{2}(\vec{b}-\vec{a})=\tfrac{1}{2}(\vec{a}+\vec{b}) \qquad\quad 2^\circ\quad \vec{a}-\tfrac{1}{2}(\vec{a}+\vec{b})=\tfrac{1}{2}(\vec{a}-\vec{b})$ $3^\circ\quad \frac{\vec{a}+\vec{b}}{2}-\frac{\vec{a}-\vec{b}}{2}=\vec{b}$.

Step 1: Common figure

Draw $\vec a=\overrightarrow{OA}$ and $\vec b=\overrightarrow{OB}$ from a common point $O$ and complete the parallelogram $OACB$, so $\overrightarrow{OC}=\vec a+\vec b$ and $\overrightarrow{BA}=\vec a-\vec b$. The diagonals $OC$ and $AB$ bisect each other at $M$, hence

\[ \overrightarrow{OM}=\tfrac12(\vec a+\vec b). \]

Step 2: Case $1^\circ$: $\vec{a}+\tfrac{1}{2}(\vec{b}-\vec{a})=\tfrac{1}{2}(\vec{a}+\vec{b})$

Analytical verification

\[ \vec{a}+\frac{1}{2}(\vec{b}-\vec{a}) = \vec{a}+\frac{1}{2}\vec{b}-\frac{1}{2}\vec{a} = \frac{1}{2}\vec{a}+\frac{1}{2}\vec{b} = \frac{1}{2}(\vec{a}+\vec{b}) \;\checkmark \]

Geometric interpretation

$\vec b-\vec a=\overrightarrow{AB}$, so $\tfrac12(\vec b-\vec a)=\overrightarrow{AM}$. Walking $O\to A\to M$ gives $\overrightarrow{OA}+\overrightarrow{AM}=\overrightarrow{OM}=\tfrac12(\vec a+\vec b)$: both sides lead from $O$ to the same point $M$.

Step 3: Case $2^\circ$: $\vec{a}-\tfrac{1}{2}(\vec{a}+\vec{b})=\tfrac{1}{2}(\vec{a}-\vec{b})$

Analytical verification

\[ \vec{a}-\frac{1}{2}(\vec{a}+\vec{b}) = \vec{a}-\frac{1}{2}\vec{a}-\frac{1}{2}\vec{b} = \frac{1}{2}\vec{a}-\frac{1}{2}\vec{b} = \frac{1}{2}(\vec{a}-\vec{b}) \;\checkmark \]

Geometric interpretation

$\vec a-\tfrac12(\vec a+\vec b)=\overrightarrow{OA}-\overrightarrow{OM}=\overrightarrow{MA}$, the vector from the midpoint $M$ to vertex $A$. It is exactly half of $\overrightarrow{BA}=\vec a-\vec b$.

Step 4: Case $3^\circ$: $\dfrac{\vec{a}+\vec{b}}{2}-\dfrac{\vec{a}-\vec{b}}{2}=\vec{b}$

Analytical verification

\[ \frac{\vec{a}+\vec{b}}{2}-\frac{\vec{a}-\vec{b}}{2} = \frac{\vec{a}+\vec{b}-\vec{a}+\vec{b}}{2} = \frac{2\vec{b}}{2} = \vec{b} \;\checkmark \]

Geometric interpretation

$\tfrac{\vec{a}+\vec{b}}{2}=\overrightarrow{OM}$, while $\tfrac{\vec{a}-\vec{b}}{2}=\overrightarrow{BM}$ (from $B$ to the midpoint $M$). Subtracting: $\overrightarrow{OM}-\overrightarrow{BM}=\overrightarrow{OM}+\overrightarrow{MB}=\overrightarrow{OB}=\vec b$.