# Thread: Vectors, ABC isoceles Triangle.

1. ## Vectors, ABC isoceles Triangle.

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.

2. ## Re: Vectors, ABC isoceles Triangle.

Originally Posted by TehNewbie
I understand Part A and realise that the length of AB = PQ
What?

Originally Posted by TehNewbie
Just unsure on how to express PQ.
Let $\lambda=\dfrac{t}{s+t}$. Show that $PB / AB = BQ/BC=\lambda$. Using $\lambda$, express vector PB through AB and BQ through BC. Then represent PQ as PB + BQ and use the fact that AB + BC = AC.

3. ## Re: Vectors, ABC isoceles Triangle.

Originally Posted by TehNewbie
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.
(b) Express the vector PQ in terms of t, s, and AC.
(c) Hence explain why PQ and AC are parallel
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.

4. ## Re: Vectors, ABC isoceles Triangle.

Originally Posted by Plato
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.
So does that mean that vector PB = (t/s+t) * vector AC ?

Edit: Yeap it sure does. Thanks for the help guys I really appreciate it

5. ## Re: Vectors, ABC isoceles Triangle.

Originally Posted by TehNewbie
So does that mean that vector PB = (t/s+t) * vector AC ?
t/(s+t), for goodness' sake!

6. ## Re: Vectors, ABC isoceles Triangle.

Originally Posted by TehNewbie
So does that mean that vector PB = (t/s+t) * vector AC ?
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 $\overrightarrow {PQ} = \left( {\frac{t}{{s + t}}} \right)\overrightarrow {AC}$

7. ## Re: Vectors, ABC isoceles Triangle.

Originally Posted by emakarov
t/(s+t), for goodness' sake!
Haha my bad incorrect notation but you get what I meant :P