# Express this vector as a linear combinations by given vectors.

Here are a parallelogram $$PQRS$$.

Let the internal dividing point of the $$\overline{PQ}$$ by $$2:1$$ is $$A$$, middle point of the $$\overline{PS}$$ is $$B$$ and intersection point between those is $$C$$

Express $$\overrightarrow {PC}$$as a linear combination of the $$\overrightarrow {PQ}$$ and $$\overrightarrow {PS}$$. Here is my attempt All I have to do find the ratio $$\alpha$$.

What should I do next? Thanks.

P.s.) Lately checking the answer sheet, it said $$\alpha$$ is $${1 \over 4}$$

• That cannot be right: $PQ$ and $RS$ are parallel so any linear combination will also be parallel to them, and $PC$ isn't. Please proof read your question carefully and edit it. – David Aug 14 at 1:29
• Thanks for your point out. I will edit it. – se-hyuck yang Aug 14 at 1:30

## 2 Answers

$$\alpha$$ can be obtained from the area ratios as follows,

$$\frac{\alpha}{1-\alpha} =\frac{\triangle BAS}{\triangle RAS}=\frac{\frac{2}{3} \cdot \frac{1}{2}\cdot\frac{1}{2}}{\frac{1}{2}}=\frac{1}{3}$$

where we observe that △BAS and △RAS share the same base AS. This allows their area ratio to be expressed as $$\alpha/(1-\alpha)$$, which is proportional to their heights. Furthermore, from the side partitions given, △RAS is $$\frac{1}{2}$$ of the area PQRS and, similarly, △BAS is $$\frac{1}{6}$$ of the area PQRS.

Thus,

$$\alpha=\frac{1}{4}$$

and $$\vec{PC}=\frac{1}{4}\vec{PQ}+\frac{5}{8}\vec{PS}$$

• Why does $$\frac{\alpha}{1-\alpha} =\frac{△BAS}{△RAS}=\frac{\frac{2}{3} \cdot \frac{1}{2}\cdot\frac{1}{2}}{\frac{1}{2}}=\frac{1}{3}$$ holds? – se-hyuck yang Aug 14 at 1:43
• Please see the added explanation in the answer. – Quanto Aug 14 at 1:49

So your work shows that $$\overrightarrow{PC}=\alpha(\overrightarrow{PQ}+\overrightarrow{PS})+\frac{1}{2}(1-\alpha)\overrightarrow{PS}$$ $$=\alpha\overrightarrow{PQ}+\frac{1}{2}(1+\alpha)\overrightarrow{PS}$$ for some $$\alpha\in(0,1)$$ since C is on $$\overline{BR}$$. By the same reasoning, since C is also on $$\overline{AS}$$ it is also $$\overrightarrow{PC}=\frac{2}{3}\beta\overrightarrow{PQ}+(1-\beta)\overrightarrow{PS}$$ for some $$\beta\in(0,1)$$. So, $$\alpha=\frac {2}{3}\beta$$ $$\frac{1}{2}(1+\alpha)=1-\beta$$ which is solved by $$\beta=\frac{6}{11}$$. So $$\overrightarrow{PC}=\frac{4}{11}\overrightarrow{PQ}+\frac{5}{11}\overrightarrow{PS}$$.