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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...

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    You can apply [Ceva's theorem](https://en.wikipedia.org/wiki/Ceva%27s_theorem). – 2017-01-26

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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 $\vec{AP}\cdot \vec{BC}-\vec{PB}\cdot\vec{CA}$?

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    Why is BC=PC-PB? – 2017-01-26
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    @SubhashisChakraborty By basic vector operation rule. – 2017-01-26
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    Ohh yeah I see it now. Let me see if I can get the rest of it now. – 2017-01-26
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    So I end up with $\vec{AP}\cdot \vec{BC}-\vec{PB}\cdot\vec{CA}$ and if I substitute for $\vec{BC} , \vec{CA}$ I end up with: $\vec{AP}\cdot ({\vec{PC}-\vec{PB}})-\vec{PB}\cdot ({\vec{PA}-\vec{PC}})$ and I can't really see how I would go about making that 0. Nothing cancels... – 2017-01-26
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    Expand, note that $\vec{AP}=-\vec{PA}$ and $\vec{AP}+\vec{PB}=\vec{AB}$. – 2017-01-26
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    I missed the second part. Thanks a lot! I really need to remember my basic vector addition... Thanks! – 2017-01-26
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    You're welcome! – 2017-01-26