# Relative to the origin O, the position vector of A is 6i+6j and the position vector of B is 12i−2j. The point C lies on vec(AB) such that vec(AC)=3/4 vec(AB). Find the position vector of C

How to find a position vector given several other position vectors with a given relationship?
Relative to the origin O , the position vector of A is $6i+6j$ and the position vector of B is $12i-2j$. The point C lies on $\stackrel{\to }{AB}$− such that $\stackrel{\to }{AC}=\frac{3}{4}\stackrel{\to }{AB}$. Find the position vector of C.
As of now, I have found $\stackrel{\to }{AB}$ and $\stackrel{\to }{AC}$:
$\stackrel{\to }{AB}=\stackrel{\to }{OB}-\stackrel{\to }{OA}$
$\stackrel{\to }{AB}=12i-2j-\left(6i+6j\right)$
$\stackrel{\to }{AB}=12i-2j-6i-6j$
$\therefore \stackrel{\to }{AB}=6i-8j$
$\stackrel{\to }{AC}=\frac{3}{4}\stackrel{\to }{AB}$
$\stackrel{\to }{AC}=\frac{3}{4}\cdot \left(6i-8j\right)$
$\stackrel{\to }{AC}=\frac{3}{4}\cdot 2\left(3i-4j\right)$
$\stackrel{\to }{AC}=\frac{3}{2}\cdot \left(3i-4j\right)$
$\stackrel{\to }{AC}=\frac{3\left(3i-4j\right)}{2}$
$\therefore \stackrel{\to }{AC}=\frac{9i-12j}{2}$
How would $\stackrel{\to }{OC}$ be found?
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Laylah Bean
You know $\stackrel{\to }{AB}=\stackrel{\to }{OB}-\stackrel{\to }{OA}$
Similarly $\stackrel{\to }{AC}=\stackrel{\to }{OC}-\stackrel{\to }{OA}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\stackrel{\to }{OC}=\stackrel{\to }{OA}+\stackrel{\to }{AC}=\left(6\stackrel{^}{i}+6\stackrel{^}{j}\right)+\left(4.5\stackrel{^}{i}-6\stackrel{^}{j}\right)=10.5\stackrel{^}{i}$