Vector Joining Two Points

Vector Joining Two Points

Vector \(\overrightarrow{AB}\) is shifted without rotation and placed at origin.

Now vector \(\overrightarrow{AB}=\overrightarrow{OC}\).

Since \(|\overrightarrow{AB}|=|\overrightarrow{OC}|\), coordinates of point C are (x₂ – x₁, y₂ – y₁, z₂ – z₁)

Hence, Vector \(\overrightarrow{OC}=\left( {{x}_{2}}-{{x}_{1}} \right)i+\left( {{y}_{2}}-{{y}_{1}} \right)j+\left( {{z}_{2}}-{{z}_{1}} \right)k\),

\(\overrightarrow{AB}=\overrightarrow{OC}=\left( {{x}_{2}}-{{x}_{1}} \right)i+\left( {{y}_{2}}-{{y}_{1}} \right)j+\left( {{z}_{2}}-{{z}_{1}} \right)k\),

\(=\overrightarrow{OB}-\overrightarrow{OA}\).

= Position vector of B – Position vector of A

Also, from above, we have \(\overrightarrow{OB}=\overrightarrow{OA}+\overrightarrow{AB}\) which describes triangle rule of vector addition.

Further \(\overrightarrow{OB}=\overrightarrow{OA}+\overrightarrow{AB}=\overrightarrow{OC}+\overrightarrow{OA}\)  \(\left( \because \ \ \overrightarrow{OC}=\overrightarrow{AB} \right)\).

Example: if \(2\overrightarrow{AC}=3\overrightarrow{CB}\), then prove that \(2\overrightarrow{OA}+3\overrightarrow{OB}=5\overrightarrow{OC}\) Where O is the origin

Solution:

Given that \(2\overrightarrow{AC}=3\overrightarrow{CB}\),

\(2\left( \overline{OC}-\overline{OA} \right)=3\left( \overline{OB}-\overline{OC} \right)\),

\(2\overrightarrow{OA}+3\overrightarrow{OB}=5\overrightarrow{OC}\).