Question:
If vectors a, b, c are position vectors of A, B, and C, with respect O respectively, and $3$OA $+ 2$OB $- 5$OC $= 0$, prove that A, B, and C are in the same straight line.
Solution:
If A, B and C are in the same straight line, OA $=$ OB $=$ OC. But that is to prove. How do I prove with only one equation given?
By simple rearrangement, we have $$\textbf{c}=\dfrac{3}{5}\textbf{a}+\dfrac{2}{5}\textbf{b}$$Consider $AC=\textbf{c}-\textbf{a}=\frac{-2}{5}(\textbf{b}-\textbf{a})$ which is parallel to $AB=(\textbf{b}-\textbf{a})$ but A is the common point, so they are colinear