A variable plane `l x+m y+n z=p(w h e r el ,m ,n` are direction cosines of normal`)` intersects the coordinate axes at points`A ,Ba n dC` , respectively. Show that the foot of the normal on the plane from the origin is the orthocenter of triangle `A B C` and hence find the coordinate of the circumcentre of triangle `A B Cdot`

Answer

`((p-l^(2)p)/(2l), (p-m^(2)p)/(2m), (p-n^(2)p)/(2n))`

Explanation

`(x)/(p//l)+(y)/(p//m)+(z)/(p//n)=1`
The foot of narmal on plane has coordinates H (lp, mp, np).
Direction ratios of AH are lp-(p/l),mp and np and direction ratios of BC are 0,-p/m and p/n. Thus,
`(lp-(p)/(l)).0+(mp)(-(p)/(m))+(np)((p)/(n))=0`

Hence, AH is perpendicular to BC.
Similarly, BH is perpendicular to AC and CH is perpendicular to AB.
Hence, H is the orthocenter.
Moreover, in any triangle, G (centroid) divides OH in the ratio 1:2. Hence,
`G-=((p)/(3l),(p)/(3m),(p)/(3n))`
H-=(lp,mp,np)
`impliesO-=((p-l^(2)p)/(2l),(p-m^(2)p)/(2m),(p-n^(2)p)/(2n))`.