Q:

altitudes of a triangle through a common point (concurence)?

So I have a triangle with the vertices $A,B$ and $C$. I'm trying to show that if the altitudes of the triangle at points A and B intersect at point $P$, then the third altitude at point $C$ must pass through $P$ as well.
I am expected to use linear algebra techniques on this but I'm a little lost on how to approach this. My guess would be to use the dot product somehow since the dot product of two perpendicular vectors is $0$ so that was my intuition but I'm not really sure how I would go about doing that...

A:

Hint: $AP\perp BC$ and $BP\perp AC$, this is equivalent to say
$$\vec{AP}\cdot \vec{BC}=0,\quad \vec{PB}\cdot\vec{CA}=0 $$
Then observe
$$\vec{BC}=\vec{PC}-\vec{PB},\quad \vec{CA}=\vec{PA}-\vec{PC}.$$
Note that to prove $CP\perp AB$, you only need to show that $\vec{CP}\cdot\vec{AB}=0$. Then what can you obtain by taking subtraction $\vec{AP}\cdot \vec{BC}-\vec{PB}\cdot\vec{CA}$?

