This is a question with regards to position vectors:
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 $\overrightarrow {AB}$ such that $\overrightarrow {AC}=\frac {3}{4}\overrightarrow {AB}$. Find the position vector of $C$.
As of now, I have found $\overrightarrow {AB}$ and $\overrightarrow {AC}$:
$\overrightarrow {AB} = \overrightarrow {OB}-\overrightarrow {OA}$
$\overrightarrow {AB} = 12i-2j-(6i+6j)$
$\overrightarrow {AB} = 12i-2j-6i-6j$
$\therefore \overrightarrow {AB} = 6i-8j$
$\overrightarrow {AC} = \frac {3}{4}\overrightarrow {AB}$
$\overrightarrow {AC} = \frac {3}{4}\cdot (6i-8j)$
$\overrightarrow {AC} = \frac {3}{4}\cdot 2(3i-4j)$
$\overrightarrow {AC} = \frac {3}{2}\cdot (3i-4j)$
$\overrightarrow {AC} = \frac {3(3i-4j)}{2}$
$\therefore \overrightarrow {AC} = \frac {9i-12j}{2}$
How would $\overrightarrow {OC}$ be found?
Looking forward to a response! Also, I'm still new to MSE, so I would really appreciate any and all feedback. Thank you!
Best Answer
You know $\vec{AB}=\vec{OB}-\vec{OA}$.
Similarly $\vec{AC}=\vec{OC}-\vec{OA}\implies\vec{OC}=\vec{OA}+\vec{AC}=(6\hat i+6\hat j)+(4.5\hat i-6\hat j)=10.5\hat i$.