Question

Using vectors prove that the altitudes of a triangle are concurrent.

Hint:

Altitudes are concurrent at the orthocenter; vector dot product conditions yield a unique intersection point.

Answer 1

Consider triangle ABC with position vectors \( \vec{A}, \vec{B}, \vec{C} \). 
Altitude from A to BC: Perpendicular to \( \vec{BC} = \vec{C} - \vec{B} \). 
Let foot be D. \( \vec{AD} \perp \vec{BC} \): \( (\vec{D} - \vec{A}) \cdot (\vec{C} - \vec{B}) = 0 \). 
\[ \vec{D} = \vec{A} + t (\vec{C} - \vec{B}), t (\vec{C} - \vec{B}) \cdot (\vec{C} - \vec{B}) = 0 \Rightarrow t = 0, \text{ but use orthocenter.} \] Orthocenter H satisfies: \( \vec{AH} \perp \vec{BC} \), \( \vec{BH} \perp \vec{CA} \), \( \vec{CH} \perp \vec{AB} \). 
\[ (\vec{H} - \vec{A}) \cdot (\vec{C} - \vec{B}) = 0, (\vec{H} - \vec{B}) \cdot (\vec{A} - \vec{C}) = 0, (\vec{H} - \vec{C}) \cdot (\vec{B} - \vec{A}) = 0. \] Solve: Let \( \vec{a} = \vec{B} - \vec{A} \), \( \vec{b} = \vec{C} - \vec{B} \), \( \vec{c} = \vec{A} - \vec{C} \). 
The equations are linear in \( \vec{H} \), and the system has a unique solution, proving concurrence. 
Answer: Proved using perpendicularity conditions.