Consider the triangle ABC. Let P divide the segment AB in the ratio s : t, and Q divide the segment BC in the ratio t : s.
(a) Draw a sketch of the situation.
(b) Express the vector PQ in terms of t, s, and AC.
(c) Hence explain why PQ and AC are parallel
I understand Part A and realise that the length of AB = PQ, and can also see how PQ is parallel to that of AC. Just unsure on how to express PQ.
I will let you make a diagram.
Note that $\overrightarrow {PQ} = \overrightarrow {PB} + \overrightarrow {BQ} $.
Now $\overrightarrow {PB} = \left( {\frac{t}{{s + t}}} \right)\overrightarrow {AB} \, \,,\;\overrightarrow {BQ} = \left( {\frac{t}{{s + t}}} \right)\overrightarrow {BC} \,\& \,\overrightarrow {AC} = \overrightarrow {AB} + \overrightarrow {BC} $
You put this together and show us what you find.
Actually from $\overrightarrow {PQ} = \overrightarrow {PB} + \overrightarrow {BQ} $.
and $\overrightarrow {PB} = \left( {\frac{t}{{s + t}}} \right)\overrightarrow {AB} \, \,,\;\overrightarrow {BQ} = \left( {\frac{t}{{s + t}}} \right)\overrightarrow {BC} \,\& \,\overrightarrow {AC} = \overrightarrow {AB} + \overrightarrow {BC} $
You get $\overrightarrow {PQ} = \left( {\frac{t}{{s + t}}} \right)\overrightarrow {AC} $