Vectors problem involving scalar multipliers.

Question

(Sorry I have had to resort to images, I am unfamiliar with vectors in mathjax)

With this question, I can happily do part (a), but it is part (b) and part (c) I am struggling with. This is my current working:

However, this is not the correct answer, any ideas? Thanks.

Answer 1

We first observe$$\overrightarrow{BC}=\overrightarrow{AC}-\overrightarrow{AB}=\frac13\overrightarrow{AB}=\binom{2}{-1}.$$For b note$$\overrightarrow{BP}=\frac12\overrightarrow{AP}=\binom{r/2}{p/2}\implies\overrightarrow{CP}=\overrightarrow{BP}-\overrightarrow{BC}=\binom{r/2-2}{p/2+1}.$$For c note that $$\overrightarrow{BP}=\overrightarrow{AP}-\overrightarrow{BP}=\overrightarrow{AB}=\binom{6}{-3}\implies\overrightarrow{CP}=\overrightarrow{BP}+\overrightarrow{AC}-\overrightarrow{AB}=\binom{8}{-4}.$$

Answer 2

Hints. You're told that the segment $AP$ is twice as long as $BP.$ You have a vector from $A$ to $P.$ You are asked for the vectors $\vec {BP}$ and $\vec {CP}.$

Note that $$\vec {AP}+\vec {PB}=\vec {AB}.$$ Since you know the first and last, you can find what you want, since it is opposite to the $\vec {PB}.$ Similarly, use the equality $$\vec {AC}+\vec {CP}=\vec {AP}$$ to find the $\vec {CP}.$ Hopefully you may now be able to do the third part (you'll likely need to use the fact that $AP=2BP$).