Question

Find the orthocentre of the triangle whose vertices are $A(a,0,0),B(0,b,0),C(0,0,c)$

Answer 1

The plane of the triangle is $x/a+y/b+z/c=1$

Let $O(\alpha ,\beta ,y)$ be the orthocenter

drs of OA are $\alpha -a,\beta ,y$

drs of BC are $o,-b,c$

$OA\perp BC\Rightarrow \beta b=yc$

$OB\perp CA\Rightarrow \alpha a=yc$

$\therefore \alpha a=\beta b=yc=k^2\left(say\right)$

$o(\alpha ,\beta ,y)$ lies on the plane

$\therefore \frac{\alpha }{a}+\frac{\beta }{b}+\frac{y}{c}=1\Rightarrow \frac{1}{k^2}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}$

The orthocentre is $(\frac{k^2}{a},\frac{k^2}{b},\frac{k^2}{c})$