Properties of Vector Addition
1. Communitive property
\(\overrightarrow{a}+\overrightarrow{b}=\overrightarrow{b}+\overrightarrow{a}\).
2. Associative property
\(\left( \overrightarrow{a}+\overrightarrow{b} \right)+\overrightarrow{c}=\overrightarrow{a}+\left( \overrightarrow{b}+\overrightarrow{c} \right)\).
3. Additive identity
\(\overrightarrow{a}+\overrightarrow{0}=\overrightarrow{a}\).
4. Additive inverse
\(\overrightarrow{a}+\left( -\overrightarrow{a} \right)=\overrightarrow{0}\).
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 \(\overrightarrow{AB},\ \overrightarrow{AE},\ \overrightarrow{BC},\ \overrightarrow{DC},\ \overrightarrow{ED}\),
\(R=\overrightarrow{AB}+\ \overrightarrow{AE}+\overrightarrow{BC}+\ \overrightarrow{DC}+\ \overrightarrow{ED}+\overrightarrow{AC}\),
\(=\left( \overrightarrow{AB}+\overrightarrow{BC} \right)+\left( \overrightarrow{AE}+\overrightarrow{BD}+\overrightarrow{DC} \right)+\left( \overrightarrow{AC} \right)\),
\(\overrightarrow{AC}+\overrightarrow{AC}+\overrightarrow{AC}=3\overrightarrow{AC}\).