Question: The median AD of the $\bigtriangleup$ABC is bisected at E, BE meets AC in F, then AF: AC is equal to.
Answer: The correct option is $\textbf{A} \frac{1}{3}$.
Taking A as the origin let the position vectors of B and C be $\overrightarrow{b}$ and $\overrightarrow{c}$ respectively.
Equations of lines BF and AC are $\overrightarrow{r} = \overrightarrow{b} + \lambda(\frac{\overrightarrow{b}+\overrightarrow{c}}{4}-\overrightarrow{b})$ and $\overrightarrow{r} = \overrightarrow{0} + \mu\overrightarrow{c}$ respectively.
For the point of intersection F, we have:
$\overrightarrow{b}+\lambda(\frac{\overrightarrow{c}-3\overrightarrow{b}}{4})=\mu \overrightarrow{c}$
$\implies 1-\frac{3\lambda}{4}=0$ and $\frac{\lambda}{4}=\mu$
$\implies \lambda=\frac{4}{3}$ and $\mu=\frac{1}{3}$
Therefore, the position vector of $\overrightarrow{F}$ is $\overrightarrow{r} = \frac{1}{3}\overrightarrow{c}$
Now, $\overrightarrow{AF} = \frac{\overrightarrow{c}}{3} \implies \overrightarrow{AF} = \frac{1}{3}\overrightarrow{AC}$
Hence, $AF : AC=\frac{1}{3}:1=\frac{1}{3}$
Yes, it is possible to have a scalar and vector part for an equation, like so: $\lambda \overrightarrow{v}, \lambda \in \mathbb{R}$