We have an equilateral triangle $ABC$ and it's height $AD$. Point $S$ lies on the line $AD$. How to express $\vec{BS}$ by $\vec{AB}$ and $\vec{BC}$. I think it's neccesery here to add a parameter but so far I haven't find it.
$\endgroup$ 11 Answer
$\begingroup$You can make use of the fact that: $$ \vec{AB} + \vec{BC} + \vec{CA} = 0 $$ to get: $$ \vec{AC} = \vec{AB} + \vec{BC} $$ Any point $S$ on the height is parametrized as: $$ \vec{AS} = \vec{AB} + \vec{BS} = \alpha\left(\frac{\vec{AB} + \vec{AC}}{2}\right) $$ where $\alpha \in[0,1]$. Thus, $$ \vec{BS} = \alpha\left(\frac{2\vec{AB} + \vec{BC} }{2}\right) - \vec{AB} \\ = \boxed{(\alpha - 1) \vec{AB} + \frac{\alpha}{2}\vec{BC}} $$ You can confirm that at the extreme values of $\alpha$, $S$ coincides with $A$ or with $D$.
$\endgroup$
