Question
Using vectors prove that the altitudes of a triangle are concurrent.
✓
Answer
In ∆ABC, Let altitudes AD and BE intersect at H.
Taking H as the origin, $\bar{a}, \bar{b}, \bar{c}$ be the position vectors of points A, B, C respectively.
Since $\overline{ HA }$ is perpendicular to $\overline{ BC }$,
$\therefore \bar{a} \cdot \overline{B C}=0$, i.e. $\bar{a} \cdot(\bar{c}-\bar{b})=0$,
i.e. $\bar{a} \cdot \bar{c}=\bar{a} \cdot \bar{b}$...(1)
Similarly $\overline{ HB }$ is perpendicular to $\overline{ AC }$.
$\therefore \bar{b} \cdot(\bar{c}-\bar{a})=0$ ,is $\bar{b} \cdot \bar{c}=\bar{a} \cdot \bar{b}$...(2)
Now we have to prove $\overline{ HC }$ is perpendicular to $\overline{ AB }$.
From (1) and (2)
$\bar{a} \cdot \bar{c}=\bar{b} \cdot \bar{c}$ i.e. $(\bar{a}-\bar{b}) \cdot \bar{c}=0$
i.e. $\overline{ BA } \cdot \overline{ HC }=0$
i.e. $\overline{ HC }$ is perpendicular to $\overline{ BA }$
Hence the altitudes of a triangle are concurrent.
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