# 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 $$2\overrightarrow{AC}=3\overrightarrow{CB}$$,

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

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