SATHEE: 3D Geometry 3 Question 65

3D Geometry 3 Question 65

####65. $T$ is a parallelopiped in which $A, B, C$ and $D$ are vertices of one face and the face just above it has corresponding vertices $A^{'}, B^{'}, C^{'}, D^{'}, T$ is now compressed to $S$ with face $A B C D$ remaining same and $A^{'}, B^{'}, C^{'}, D^{'}$ shifted to $A^{''}, B^{''}, C^{''}, D^{''}$ in $S$ . The volume of parallelopiped $S$ is reduced to $90$ of $T$ . Prove that locus of $A^{''}$ is a plane.

$(2004, 2 M)$

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Solution:

Let the equation of the plane $A B C D$ be $a x + b y + c z + d = 0$ , the point $A^{''}$ be $(α, β, γ)$ and the height of the parallelopiped $A B C D$ be $h$ .

$\begin{array}{lrl} \Rightarrow & \frac{| a α + b β + c γ + d |}{\sqrt{a^{2} + b^{2} + c^{2}}} = 90 \Rightarrow & a α + b β + c γ + d = \pm 0.9 h \sqrt{a^{2} + b^{2} + c^{2}} \end{array}$

$∴$ Locus is $a x + b y + c z + d = \pm 0.9 h \sqrt{a^{2} + b^{2} + c^{2}}$

Hence, locus of $A^{''}$ is a plane parallel to the plane $A B C D$ .

3D Geometry 3 Question 65

Solution:

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3D Geometry 3 Question 65

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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