A parallelepiped `S` has base points `A ,B ,Ca n dD` and upper face points `A^(prime),B^(prime),C^(prime),a n dD '` . The parallelepiped is compressed by upper face `A ' B ' C ' D '` to form a new parallepiped `T` having upper face points `A^(prime prime),B^(prime prime),C^(prime prime) and D^(prime prime)` . The volume of parallelepiped `T` is 90 percent of the volume of parallelepiped `Sdot` Prove that the locus of `A^(prime prime)` is a plane.

Explanation

S is the parallelepiped with base point `A, B, C and D` and upper face points `A', B', C' and D'`. Let its volume ve `V_(s)`. By compressing it by upper face `A', B', C' and D'`, a new parallelepiped T is formed whose upper face points are now `A'', B'', C'' and D''`. Let its volume be `V_(T)`.
Let `h` be the height of original parallelepiped S. Then
`" "V_(S)= ("area of " ABCD)xxh" "` (i)
Let equation of plane `ABCD` be `ax+by +cz+d=0 and A''(alpha, beta, gamma)`.

Then the height of the new parallelepiped T is the length of the perpendicular from `A''` to `ABCD`, i.e., `(aalpha +bbeta+cgamma+d)/(sqrt(a^(2)+b^(2)+c^(2))`. Therefore
`" "V_(T)= ("ar "ABCD)xx((aalpha+bbeta+cgamma+d))/(sqrt(a^(2)+b^(2)+c^(2))" "` (ii)
But given that `V_(T)= (90)/(100)V_(s)" "` (iii)
From (i), (ii) and (iii), we get
`" "(aalpha+bbeta+cgamma+d)/(sqrt(a^(2)+b^(2)+c^(2))= 0.9 h`
or `" "aalpha+bbeta+cgamma+d-0.9 hsqrt(a^(2)+b^(2)+c^(2))=0`
Therefore, the locus of `A'' (alpha, beta, gamma)` is `ax+by +cz+d -0.9 h sqrt(a^(2)+b^(2)+c^(2))=0`, which is a plane parallel to `ABCD`. Hence proved.