1. Communitive property

$$\overrightarrow{a}+\overrightarrow{b}=\overrightarrow{b}+\overrightarrow{a}$$.

2. Associative property

$$\left( \overrightarrow{a}+\overrightarrow{b} \right)+\overrightarrow{c}=\overrightarrow{a}+\left( \overrightarrow{b}+\overrightarrow{c} \right)$$.

$$\overrightarrow{a}+\overrightarrow{0}=\overrightarrow{a}$$.

$$\overrightarrow{a}+\left( -\overrightarrow{a} \right)=\overrightarrow{0}$$.

5. $$\left| \overrightarrow{a}+\overrightarrow{b} \right|\le \left| \overrightarrow{a} \right|+\left| \overrightarrow{b} \right|$$and $$\left| \overrightarrow{a}-\overrightarrow{b} \right|\ge \left| \overrightarrow{a} \right|-\left| \overrightarrow{b} \right|$$.

Examples 1: If vector $$\overrightarrow{a}+\overrightarrow{b}$$ bisects the angle between $$\overrightarrow{a}$$ and $$\overrightarrow{b}$$, then prove that $$\left| \overrightarrow{a} \right|=\left| \overrightarrow{b} \right|$$.

Solution: We know that vector $$\overrightarrow{a}+\overrightarrow{b}$$ is along the diagonal of the parallelogram whose adjacent sides are vectors $$\overrightarrow{a}$$ and $$\overrightarrow{b}$$.

Now if $$\overrightarrow{a}+\overrightarrow{b}$$ bisects the angle between vector $$\overrightarrow{a}$$ and $$\overrightarrow{b}$$, then the parallelogram must be a rhombus

Hence  $$\left| \overrightarrow{a} \right|=\left| \overrightarrow{b} \right|$$.

Example 2: $$\overrightarrow{AO}+\overrightarrow{OB}=\overrightarrow{BO}+\overrightarrow{OC}$$, then prove that B is the midpoint of AC.

Solution: Given that $$\overrightarrow{AO}+\overrightarrow{OB}=\overrightarrow{BO}+\overrightarrow{OC}$$,

$$\overrightarrow{AB}=\overrightarrow{BC}$$,

Thus, vectors $$\overrightarrow{AB}$$ and $$\overrightarrow{BC}$$ are collinear

Point A, B, C are collinear

$$\overrightarrow{AB}=\overrightarrow{BC}$$,

B is the midpoint of AC.

Example 3: ABCDE is a pentagon prove that the resultant of forces $$\overrightarrow{AB},\ \overrightarrow{AE},\ \overrightarrow{BC},\ \overrightarrow{DC},\ \overrightarrow{ED}$$ and $$\overrightarrow{AC}$$ is $$3\overrightarrow{AC}$$.

Solution: Given that $$\overrightarrow{AB},\ \overrightarrow{AE},\ \overrightarrow{BC},\ \overrightarrow{DC},\ \overrightarrow{ED}$$,

$$R=\overrightarrow{AB}+\ \overrightarrow{AE}+\overrightarrow{BC}+\ \overrightarrow{DC}+\ \overrightarrow{ED}+\overrightarrow{AC}$$,

$$=\left( \overrightarrow{AB}+\overrightarrow{BC} \right)+\left( \overrightarrow{AE}+\overrightarrow{BD}+\overrightarrow{DC} \right)+\left( \overrightarrow{AC} \right)$$,

$$\overrightarrow{AC}+\overrightarrow{AC}+\overrightarrow{AC}=3\overrightarrow{AC}$$.