3D Geometry Ques 75
75. $T$ is a parallelopiped in which $A, B, C$ and $D$ are vertices of one face and the face just above it has corresponding vertices $A^{\prime}, B^{\prime}, C^{\prime}, D^{\prime}, T$ is now compressed to $S$ with face $A B C D$ remaining same and $A^{\prime}, B^{\prime}, C^{\prime}, D^{\prime}$ shifted to $A^{\prime \prime}, B^{\prime \prime}, C^{\prime \prime}, D^{\prime \prime}$ in $S$. The volume of parallelopiped $S$ is reduced to $90 %$ of $T$. Prove that locus of $A^{\prime \prime}$ is a plane.
$(2004,2 \mathrm{M})$
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Solution:
Formula:
Let the equation of the plane $A B C D$ be $a x+b y+c z+d=0$, the point $A^{\prime}$ be $(\alpha, \beta, \gamma)$ and the height of the parallelepiped $A B C D$ be $h$.
$ \begin{array}{lrl} \Rightarrow & \frac{|a \alpha+b \beta+c \gamma+d|}{\sqrt{a^{2}+b^{2}+c^{2}}}=1 \% \quad h \\ & \\ \Rightarrow & a \alpha + b \beta + c \gamma + d = \pm 0.9 h \sqrt{a^{2}+b^{2}+c^{2}} \end{array} $
$\therefore \quad$ Locus is $a x+b y+c z+d= \pm h \sqrt{a^{2}+b^{2}+c^{2}}$
Hence, locus of $A^{\prime \prime}$ is a plane parallel to the plane $A B C D$.
BETA
