Pure mathematics: parallel vectors using ratios

Question

The question is as follows:

In $\triangle OAB$, $\vec{OA}=\mathbf a$ and $\vec{OB}=\mathbf b$. $P$ divides $OA$ in the ratio $3:2$ and $Q$ divides $OB$ in the ratio $3:2$

1. Show that $PQ$ is parallel to $AB$.
2. Given that $AB$ is $10\,\mathrm{cm}$ in length find the length of $PQ$.

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I don't understand how to do the first part. I think you have to generate a scale factor.

Answer 1

You could say, using vector calculus that:

Let us call $OA$ for vector $\bar{a}=(a_1,a_2)$ and $OB$ for vector $\bar{b}=(b_1,b_2)$. Then the line segment $AB$, let us call it $\bar{x}$ is just $\bar{a}-\bar{b}=(a_1-b_1,a_2-b_2)$. Similar way you can see that $OP$ and $OQ$ can also be described as vectors, namely $\frac{3}{5}\bar{a}$ and $\frac{3}{5}\bar{b}$. The vector $\bar{y}$ pointing from $Q$ to $P$ is the difference between $\frac{3}{5}\bar{a}$ and $\frac{3}{5}\bar{b}$ that is $$ \bar{y}=\Big(\frac{3}{5}a_1-\frac{3}{5}b_1,\frac{3}{5}a_2-\frac{3}{5}b_2\Big)=\frac{3}{5}(a_1-b_1,a_2-b_2). $$ And as we can see $\bar{x}$ is just a scaled up version of $\bar{y}$ and therefore they are parrallel.

Answer 2

You can use side angle side similarity criterion to prove that $\triangle PQO$ and $\triangle ABO$ are similar. Thus $\angle OPQ = \angle OAB$ which means that lines are parallel as the corresponding angles are equal.