JEE Advanced 2004

LEVELJEE Main

Animated Solution for Mathematics - Three Dimensional Geometry: A parallelopiped 'S' has base points $A, B, C$ and $D$ and upper face points $A^{'}, B^{'}, C^{'}$ and $D^{'}$ . This parallelopiped is compressed by upper face $A^{'} B^{'} C^{'} D^{'}$ to form a new parallelopiped 'T' having upper face points $A^{''}, B^{''}, C^{''}$ and $D^{''}$ . Volume of parallelopiped $T$ is 90 percent of the volume of parallelopiped $S$ . Prove that the locus of $A^{''}$ , is a plane.

Q: Why is the locus a plane? Shouldn't it be a point or a line?

Think about the degrees of freedom. The point $A''$ is a vertex of the top face. As the top face is compressed, $A''$ can move within the plane of the new top face. Since the height $h'$ is constant for all points on that top face, the locus of all such points $A''$ that maintain this specific height relative to the base plane must form a surface parallel to the base. In 3D space, that surface is a plane.

Q: What about the $\pm$ sign when we remove the absolute value? Does it matter?

Great catch! When we remove the absolute value $|ax + by + cz + d| = k$, we get $ax + by + cz + d = \pm k$. This mathematically represents two possible planes, one on either side of the base plane. In the context of our physical compression, we are moving the top face towards the base, so we choose the sign that corresponds to the region where the parallelepiped exists. Both represent valid parallel planes.

Progressive Hints

Hint 1

Hint 1 (The Spark): Visualize the parallelepiped as a stack of layers. Since the base

A B C D

is fixed, the volume of the parallelepiped is directly proportional to its height

h

. When you compress the top face, you are essentially performing a vertical scaling. The key is to realize that the base area

A_{ba se}

remains invariant throughout this transformation.

Hint 2

Hint 2 (The Arsenal): Recall the fundamental 3D geometry formula for the perpendicular distance

h

from a point

P (α, β, γ)

to a plane

a x + b y + cz + d = 0

, which is

h = \frac{∣ a α + b β + c γ + d ∣}{a ^{2} + b ^{2} + c ^{2}}

. This formula is your bridge between the algebraic coordinates of the top face and the geometric height of the parallelepiped.

Hint 3

Hint 3 (The Execution): Start by establishing the relationship between the new volume

V_{T}

and the original volume

V_{S}

V_{T} = 0.9 V_{S}

. Since

V = A_{ba se} \times h

, this implies

h^{'} = 0.9 h

. Now, set the distance formula for the new top face point

A^{''} (α, β, γ)

equal to

0.9 h

and solve for the locus of

(α, β, γ)

Visualized Solution

Visualizing Parallelepiped $S$

Consider parallelepiped $S$ with base $A B C D$ and top face $A^{'} B^{'} C^{'} D^{'}$ .
Let the area of the base $A B C D$ be $A_{ba se}$ .
Let the height of $S$ be $h$ .

Volume of Parallelepiped $S$

The volume of a parallelepiped is the product of its base area and height.
Volume = Area of base $\times$ Height
For $S$ : $V_{S} = A_{ba se} \times h$

Defining the Base Plane

Let the equation of the base plane containing $A B C D$ be:
Plane Equation: $a x + b y + cz + d = 0$

Compression to Parallelepiped $T$

Parallelepiped $S$ is compressed to form a new parallelepiped $T$ .
The base $A B C D$ remains fixed.
The new upper face is $A^{''} B^{''} C^{''} D^{''}$ .

Volume Relation

We are given that the volume of $T$ is $90%$ of the volume of $S$ .
Relation: $V_{T} = 0.9 V_{S}$

Finding the New Height $h^{'}$

Let the height of $T$ be $h^{'}$ .
$V_{T} = A_{ba se} \times h^{'}$
Substituting into the relation: $A_{ba se} \times h^{'} = 0.9 (A_{ba se} \times h)$
New Height: $h^{'} = 0.9 h$

Distance Formula for $A^{''}$

Let the coordinates of $A^{''}$ be $(α, β, γ)$ .
The height $h^{'}$ is the perpendicular distance from $A^{''}$ to the base plane $a x + b y + cz + d = 0$ .
Distance Formula: $h^{'} = \frac{∣ a α + b β + c γ + d ∣}{a ^{2} + b ^{2} + c ^{2}}$

Equating the Heights

We know $h^{'} = 0.9 h$ .
Equating the two expressions for $h^{'}$ :
$\frac{∣ a α + b β + c γ + d ∣}{a ^{2} + b ^{2} + c ^{2}} = 0.9 h$

Finding the Locus Equation

Remove the absolute value by introducing $\pm$ :
$a α + b β + c γ + d = \pm 0.9 h a^{2} + b^{2} + c^{2}$
To find the locus, replace $(α, β, γ)$ with general coordinates $(x, y, z)$ :
Locus: $a x + b y + cz + (d \mp 0.9 h a^{2} + b^{2} + c^{2}) = 0$

Conclusion: Parallel Plane

The locus equation is of the form $A x + B y + C z + D^{'} = 0$ .
Since the coefficients $a, b, c$ are identical to the base plane, the normal vectors are parallel.
Conclusion: The locus of $A^{''}$ is a plane parallel to the base plane $A B C D$ .

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The Sigma Insight: Equation of a Plane

Solution Diagram

Analyzing the Setup

Imagine you are standing in a vast, empty 3D coordinate space. Before you sits a perfect, rigid parallelepiped

S

. It is anchored to the ground by its base,

A B C D

You can see the top face,

A^{'} B^{'} C^{'} D^{'}

, hovering at a height

h

above the base. The volume of this parallelepiped is the product of its base area,

A_{ba se}

, and its perpendicular height,

h

That is, the volume is defined as:

V_{S} = A_{ba se} \times h

The Transformation

Now, imagine applying a force to the top face

A^{'} B^{'} C^{'} D^{'}

. We are compressing the parallelepiped

S

into a new shape,

T

. The base

A B C D

remains perfectly fixed as our reference frame.

We are told that the volume of this new parallelepiped

T

90%

of the original volume

S

. Mathematically, we express this as:

V_{T} = 0.9 V_{S}

Since the base area

A_{ba se}

is constant, the change in volume must be entirely due to a change in height. Let the new height be

h^{'}

. Our volume equation becomes:

A_{ba se} \times h^{'} = 0.9 (A_{ba se} \times h)

Because

A_{ba se}

is non-zero, we can elegantly cancel it from both sides. This reveals that the new height is simply:

h^{'} = 0.9 h

Connecting Algebra to Geometry

To find the locus of the point

A^{''}

, let's assign it coordinates

(α, β, γ)

. We know that the base plane can be represented by the general equation

a x + b y + cz + d = 0

The perpendicular distance from any point

(α, β, γ)

to this plane is given by the classic distance formula:

h^{'} = \frac{∣ a α + b β + c γ + d ∣}{a ^{2} + b ^{2} + c ^{2}}

We already know that

h^{'} = 0.9 h

. By equating these two expressions, we get:

\frac{∣ a α + b β + c γ + d ∣}{a ^{2} + b ^{2} + c ^{2}} = 0.9 h

The Final Revelation

To clear the absolute value, we introduce the

\pm

sign. We then replace the specific coordinates

(α, β, γ)

with the general variables

(x, y, z)

to define the locus:

a x + b y + cz + d = \pm 0.9 h a^{2} + b^{2} + c^{2}

Rearranging this, we obtain the equation for the locus:

a x + b y + cz + (d \mp 0.9 h a^{2} + b^{2} + c^{2}) = 0

This is a linear equation in

x, y,

and

z

. In 3D space, any equation of the form

A x + B y + C z + D^{'} = 0

represents a plane.

Furthermore, the coefficients of

x, y,

and

z

are

a, b,

and

c

—the exact same coefficients as our base plane. This means the normal vector of our new locus is identical to the normal vector of the base plane.

The locus of $A^{''}$ is a plane perfectly parallel to the base plane $A B C D$ .

Common Traps & FAQs

Wait, why do we assume the base area

A_{ba se}

is constant? Doesn't compression change the base?

That is a fantastic observation! In this specific problem, the compression is defined as acting on the top face

A^{'} B^{'} C^{'} D^{'}

. The base

A B C D

is explicitly stated to remain fixed. Since the base area of a parallelepiped is determined solely by the vectors forming the base, and those vectors are unchanged, the base area remains constant. We are only changing the height.

Why is the locus a plane? Shouldn't it be a point or a line?

Think about the degrees of freedom. The point

A^{''}

is a vertex of the top face. As the top face is compressed,

A^{''}

can move within the plane of the new top face. Since the height

h^{'}

is constant for all points on that top face, the locus of all such points

A^{''}

that maintain this specific height relative to the base plane must form a surface parallel to the base. In 3D space, that surface is a plane.

What about the

\pm

sign when we remove the absolute value? Does it matter?

Great catch! When we remove the absolute value

∣ a x + b y + cz + d ∣ = k

, we get

a x + b y + cz + d = \pm k

. This mathematically represents two possible planes, one on either side of the base plane. In the context of our physical compression, we are moving the top face towards the base, so we choose the sign that corresponds to the region where the parallelepiped exists. Both represent valid parallel planes.

Visualized Solution

Visualizing Parallelepiped $S$

Volume of Parallelepiped $S$

Defining the Base Plane

Compression to Parallelepiped $T$

Volume Relation

Finding the New Height $h^{'}$

Distance Formula for $A^{''}$

Equating the Heights

Finding the Locus Equation

Conclusion: Parallel Plane

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Transformation

Connecting Algebra to Geometry

The Final Revelation

Similar Questions

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Visualized Solution

Visualizing Parallelepiped .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

Volume of Parallelepiped .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

Defining the Base Plane

Compression to Parallelepiped .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T

Volume Relation

Finding the New Height .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h′

Distance Formula for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A′′

Equating the Heights

Finding the Locus Equation

Conclusion: Parallel Plane

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Transformation

Connecting Algebra to Geometry

The Final Revelation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Two system of rectangular axes have the same origin. If a plane cuts them at distances a,b,c and a′,b′,c′ from the origin then

Visualizing Parallelepiped $S$

Volume of Parallelepiped $S$

Compression to Parallelepiped $T$

Finding the New Height $h^{'}$

Distance Formula for $A^{''}$

Two system of rectangular axes have the same origin. If a plane cuts them at distances $a, b, c$ and $a^{'}, b^{'}, c^{'}$ from the origin then