Suppose that O, A and B are three non-collinear points in a plane Let OC=OB-2OA, OD=OB+3OA, OE=-OA
I have found that OM is equal to 1/3(OB-2OA), ON is equal to 1/4(OB+3OA) and OP is equal to 2/7(OB - 3OA) + OA
Determine if the point P lies on the line through M and N.
How do I go about answering this question? Step by step answers would be preferred :)
Try picturing a triangle, with vertices marked O, M, N respectively, so that
$$\overrightarrow{MN}=\overrightarrow{ON}-\overrightarrow{OM}$$
If P lies on the line through M and N, for some real number $\lambda$, it must hold that $$\overrightarrow{OP}+\lambda\overrightarrow{MN}=\overrightarrow{OM} (\ or\ \overrightarrow{ON})$$
In this plane, all vectors can be expressed by a linear combination of two non-collinear vectors. For each pair of two non-collinear vectors to represent a particular vector, the combination is unique. Pick $\overrightarrow{OA}\ and\ \overrightarrow{OB}$ for convenience.
$$\overrightarrow{OM}=\frac{1}{3}\overrightarrow{OB}-\frac{2}{3}\overrightarrow{OA}$$ $$\overrightarrow{ON}=\frac{1}{4}\overrightarrow{OB}+\frac{3}{4}\overrightarrow{OA}$$ So $$\overrightarrow{MN}=\overrightarrow{ON}-\overrightarrow{OM}=-\frac{1}{12}\overrightarrow{OB}+\frac{17}{12}\overrightarrow{OA}$$ $$\overrightarrow{OP}=\frac{2}{7}\overrightarrow{OB}+\frac{1}{7}\overrightarrow{OA}$$ Solve the equations (by implementing $\overrightarrow{OP}+\lambda\overrightarrow{MN}=\overrightarrow{OM}$) : $$ \frac{2}{7}-\frac{1}{12}\lambda=\frac{1}{3}$$ $$ \frac{1}{7}+\frac{17}{12}\lambda=-\frac{2}{3}$$ The solutions for both equations should be the same if P lies on the line through M and N, otherwise point M, N, and P are non-collinear. In this case, $\lambda=-\frac{4}{7}$ for both equation, so it does lie on the line through M and N. With this $\lambda=-\frac{4}{7}$, it is also easy for you to point out the exact position of P. Hope this might help :)