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A parallelepiped S has base points A ,B ...

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",B",C"a n dD"` . The volume of parallelepiped `T` is 90 percent of the volume of parallelepiped `Sdot` Prove that the locus of `A"` is a plane.

Text Solution

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Let the equation of the plane ABCD be `ax+by+cz+d=0,` the point A''be `(alpha,beta,gamma)` and the height of the parallelopiped ABCD be h.
`implies" "(|aa+b beta+cy+d|)/(sqrt(a^(2)+b^(2)+c^(2)))=90%h`
`implies" "aa+b beta+cy+d=+-0.9hsqrt(a^(2)+b^(2)+c^(2))`
`:.` Locus is `ax+by+cz+d=+-0.9hsqrt(a^(2)+b^(2)+c^(2))`
Hence, locus of A'' is a plane parallel to the plane ABCD.
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