Averages and Vector Arithmetic Explanation

Question

I had a question on how they derived this equation below: I don't see the equality or know how to prove the equality below:

Detailed: I don't see how the left side of the equation involving vector OM and the right side of the equation are equal. They didn't even put the thing on some metric space or space with an axis so I can't judge why half of the length of the vector describing the length from $P_1$ to $P_2$ is equal to half of the length of vector O$P_2$ minus O$P_1$. Can I get an explanation on the equality of the two sides of how vector O$P_1$ plus one-half the vector from $P_1$ to $P_2$ equals vector O$P_1$ plus one-half of the quantity of the vector O$P_2$ minus the vector O$P_1$?

Answer 1

The following assumes that $\,M\,$ is the midpoint of $\,P_1P_2\,$, which was not explicitly stated in the question, but is implied by the context. In that case, $\displaystyle\,\overrightarrow{P_1M}=\frac{1}{2}\,\overrightarrow{P_1P_2}\,$.

$\displaystyle\overrightarrow{OM}=\overrightarrow{OP_1}+\frac{1}{2}\,\overrightarrow{P_1P_2}=\dots$

Since $\displaystyle\,\frac{1}{2}\,\overrightarrow{P_1P_2}=\overrightarrow{P_1M}\,$, the above follows from $\,\overrightarrow{OM}=\overrightarrow{OP_1}+\overrightarrow{P_1M}\,$ by the vector addition definition, sometimes referred to in this context as the head-to-tail rule.

$\displaystyle\dots=\overrightarrow{OP_1}+\frac{1}{2}\left(\overrightarrow{OP_2} - \overrightarrow{OP_1}\right)\,$

This follows from the previous step because $\,\overrightarrow{P_1P_2}=\overrightarrow{OP_2}-\overrightarrow{OP_1} \iff \overrightarrow{OP_1}+\overrightarrow{P_1P_2}=\overrightarrow{OP_2}\,$, which in turn follows from the same vector addition property as above.