If `P` is any point on the plane `l x+m y+n z=pa n dQ` is a point on the line `O P` such that `O P.O Q=p^2` , then find the locus of the point `Qdot`

Answer

`p(lx+my+nz)=(x^(2) + y^(2)+ z^(2))`

Explanation

Let `P(alpha,beta,gamma)and(x_(1),y_(1),z_(1))` be the given points. Direction ratios of OP are `alpha,betaandgamma` and those of OQ are `x_(1),y_(1)andz_(1)`
Since O,Q and P are collinear, we have
`(alpha)/(zx_(1))=(beta)/(y_(1))=(gamma)/(z_(1))=k" "(say)" "(i)`
AS `P(alpha,beta,gamma)` lies on the plane `lx+my+nz=p,lalpha+mbeta+ngamma=p`
or `klx_(1)+kmy_(1)+knz_(1)=0" "["using(i)"]" "(ii)`
Since `OP.OQ=p^(2)`, we have
`sqrt(alpha^(2)+beta^(2)+gamma^(2)).sqrt(x_(1)^(2)+y_(1)^(2)+z_(1)^(2))=p^(2)`
or `sqrt(k^(2)x_(1)^(2)+k^(2)y_(1)^(2)+k^(2)z_(1)^(2)).sqrt(x_(1)^(2)+y_(1)^(2)+z_(1)^(2))=p^(2)`
or `k(x_(1)^(2)+y_(1)^(2)+z_(1)^(2))=p^(2)" "(iii)`
From (ii) and (iii)
`(lx_(1)+my_(1)+nz_(1))/(x_(1)^(2)+y_(1)^(2)+z_(1)^(2))=(1)/(p)`
or `p(lx_(1)+my_(1)+nz_(1))=(x_(1)^(2)+y_(1)^(2)+z_(1)^(2))`
Hence, the locus of Q is `p(lx_(1)+my_(1)+nz_(1))=(x^(2)+y^(2)+z^(2))`