Question

A variable plane which remains at a constant distance 3p from the origin cuts the coordinate axes at A, B, C. Show that the locus of the centroid of triangle ABC is `1/x^2+1/y^2+1/z^2=1/p^2`.

Answer 1

Let the equation of the variable plane be
`x/a+y/b+z/c=1`………….(i)
This plane meets the x-axis, and z-axis at the points A(a,0,0), B(0,b,0)` and `C(0,0,c)` respectively. Let `(alpha,beta,gamma)` be the coordinates of the centroid of `triangleABC`.
Then, `alpha=(a+0+0)/(3), beta=(0+b+0)/(3)` and `gamma=(0+0+c)/3`
`rArr alpha=a/3, beta=b/3` and `gamma=c/3 rArr a=3alpha, b=3beta` and `c=3gamma`..........(iii)
`therefore 3p`= length of the perpendicular from (0,0,0) to the plane (i)
`rArr 3p=|0/a+0/b+0/c-1|/sqrt(1/a^(2)+1/b^(2)+1/c^(2))=1/sqrt(1/a^(2)+1/b^(2)+1/c^(2))`
`rArr sqrt(1/a^(2)+1/b^(2)+1/c^(2))=1/(3p)`
`rArr 1/a^(2)+1/b^(2)+1/c^(2)=1/(9p^(2)) rArr 1/(9alpha^(2))+1/(9beta^(2))+1/(9gamma^(2))=1/(9p^(2))` [using (ii)]
`rArr alpha^(-2)+beta^(-2)+gamma^(-2)=p^(-2)`.
Hence, the required locus is `x^(-2)+y^(-2)+z^(-2)=p^(-2)`.