JEE Advanced 1983

LEVELBoard

Animated Solution for Mathematics - Vector Algebra: The volume of the parallelopiped whose sides are given by $O A = 2 \hat{i} - 2 \hat{j}, O B = \hat{i} + \hat{j} - \hat{k}, O C = 3 \hat{i} - \hat{k}$ , is

Select Answer:

Progressive Hints

Hint 1

Hint 1: Visualize the parallelepiped as a 3D box defined by three vectors originating from the same point. The volume is not just a simple product of lengths; it is the product of the base area and the perpendicular height. Think about how the cross product of two vectors creates a vector representing the area and orientation of the base.

Hint 2

Hint 2: To calculate this volume, you need the Scalar Triple Product, often called the box product. This is defined as the dot product of one vector with the cross product of the other two:

V = ∣ O A \cdot (O B \times O C) ∣

. This operation is elegantly computed using the determinant of a

3 \times 3

matrix formed by the components of the three vectors.

Hint 3

Hint 3: Set up your determinant using the components of the vectors as rows:

D = det 213 - 2 10 0 - 1 - 1

. Expand this determinant along the first row to find the scalar triple product. Remember, the volume is the absolute value of this result.

Visualized Solution

Visualizing the Vectors $O A$ , $O B$ , and $O C$

We are given three vectors representing the adjacent edges of a parallelepiped.
Let the origin be $O (0, 0, 0)$ .
The first edge is $O A = 2 \hat{i} - 2 \hat{j} + 0 \hat{k}$ .
The second edge is $O B = 1 \hat{i} + 1 \hat{j} - 1 \hat{k}$ .
The third edge is $O C = 3 \hat{i} + 0 \hat{j} - 1 \hat{k}$ .

Constructing the Parallelepiped

By translating these vectors, we construct a 3D parallelepiped.
The solid has $6$ faces, each of which is a parallelogram.
The volume is a measure of the total 3D space enclosed by these faces.

The Scalar Triple Product

The volume $V$ of a parallelepiped is given by the absolute value of the Scalar Triple Product.
Formula: $V = ∣ [O A O B O C] ∣$
This is mathematically equivalent to: $V = ∣ O A \cdot (O B \times O C) ∣$

Setting up the Determinant

We write the components of the vectors as rows of a $3 \times 3$ matrix.
$O A = (2, - 2, 0)$
$O B = (1, 1, - 1)$
$O C = (3, 0, - 1)$
The determinant is: $D = det 213 - 2 10 0 - 1 - 1$

Expanding along the First Row

Expanding along Row 1:
$D = a_{11} C_{11} + a_{12} C_{12} + a_{13} C_{13}$
$D = 2 det (10 - 1 - 1) - (- 2) det (13 - 1 - 1) + 0$

Calculating the First Minor

First term: $2 det (10 - 1 - 1)$
Evaluate the $2 \times 2$ determinant:
$det (10 - 1 - 1) = (1 \times (- 1)) - ((- 1) \times 0)$
$= - 1 - 0 = - 1$
First term value: $2 \times (- 1) = - 2$

Calculating the Second Minor

Second term: $- (- 2) det (13 - 1 - 1) = + 2 det (13 - 1 - 1)$
Evaluate the $2 \times 2$ determinant:
$det (13 - 1 - 1) = (1 \times (- 1)) - ((- 1) \times 3)$
$= - 1 - (- 3) = - 1 + 3 = 2$
Second term value: $2 \times 2 = 4$

Combining the Terms

Combine the calculated values of the terms:
$D = (First Term) + (Second Term) + 0$
$D = - 2 + 4$
$D = 2$

Comparing with Options

The volume of the parallelepiped is $V = ∣ D ∣ = 2$ cubic units.
Let's check the given options:
Option A: $\frac{4}{13}$
Option B: $4$
Option C: $\frac{2}{7}$
Since $2$ is not present in A, B, or C, the correct option is none of these.

00:00 / 00:00

The Sigma Insight: Scalar Triple Product

Solution Diagram

Analyzing the Setup

We are tasked with finding the volume of a parallelepiped defined by three vectors,

O A

O B

, and

O C

, originating from the origin

O

. These vectors serve as the adjacent edges of the solid.

The volume

V

of a parallelepiped formed by three vectors is given by the absolute value of their Scalar Triple Product. This is mathematically expressed as:

V = ∣ O A \cdot (O B \times O C) ∣

The Master Tool

Scalar Triple Product

The cross product

O B \times O C

yields a vector whose magnitude represents the area of the base parallelogram and whose direction is normal to that base. By taking the dot product of

O A

with this normal vector, we effectively multiply the base area by the perpendicular height of the solid.

This geometric interpretation confirms that the scalar triple product is the most efficient way to calculate the volume of a parallelepiped.

The Computational Engine

Determinants

To compute this efficiently, we represent the vectors as rows in a

3 \times 3

matrix. The volume is the absolute value of the determinant of this matrix:

D = det 213 - 2 10 0 - 1 - 1

We expand this determinant along the first row:

D = 2 10 - 1 - 1 - (- 2) 13 - 1 - 1 + 01310

Calculating the minors:

1. The first term:

2 \times ((1 \times - 1) - (- 1 \times 0)) = 2 \times (- 1) = - 2

. 2. The second term:

2 \times ((1 \times - 1) - (- 1 \times 3)) = 2 \times (- 1 + 3) = 2 \times 2 = 4

. 3. The third term is

0

Summing these values, we find

D = - 2 + 4 = 2

. Thus, the volume is

∣2∣ = 2

cubic units.

The 'None of These' Trap

We have calculated the volume to be

2

cubic units. Comparing this to the provided options (

4/13

4

2/7

, and 'none of these'), we see that our result is not explicitly listed.

In JEE Advanced examinations, encountering a result that does not match the provided options can be unsettling. However, if your derivation follows the correct geometric principles and your arithmetic is verified, you must trust your result.

The correct choice is 'none of these'. This serves as a reminder to maintain confidence in your mathematical process and to remain focused under pressure.

Common Traps & FAQs

Wait, why do we take the absolute value of the determinant? Can volume be negative?

That is a fantastic question! In pure mathematics, the scalar triple product can indeed be negative depending on the orientation of the vectors (the 'handedness' of the coordinate system). However, in physics and geometry, volume represents a physical quantity of space, which must always be non-negative. Taking the absolute value ensures our final answer is a physically meaningful, positive magnitude.

Does the order of the vectors in the determinant change the volume?

Changing the order of the rows in a determinant only changes its sign (multiplying by

- 1

). Since the volume of a parallelepiped is the absolute value of the scalar triple product, changing the order of the vectors will not change the final volume. You can place the vectors in any row order you like, and you will arrive at the same magnitude.

What does it mean if the determinant equals zero?

If the determinant is zero, it means the scalar triple product is zero. Geometrically, this implies that the three vectors are coplanar—they all lie in the same flat plane. If they are coplanar, they cannot enclose any three-dimensional space, so the volume of the 'parallelepiped' is zero. It collapses into a flat shape!

Animated Solution for Mathematics - Vector Algebra: The volume of the parallelopiped whose sides are given by $O A = 2 \hat{i} - 2 \hat{j}, O B = \hat{i} + \hat{j} - \hat{k}, O C = 3 \hat{i} - \hat{k}$ , is

Select Answer:

Visualized Solution

Visualizing the Vectors $O A$ , $O B$ , and $O C$

Constructing the Parallelepiped

The Scalar Triple Product

Setting up the Determinant

Expanding along the First Row

Calculating the First Minor

Calculating the Second Minor

Combining the Terms

Comparing with Options

The Sigma Insight: Scalar Triple Product

Analyzing the Setup

The Master Tool

The Computational Engine

The 'None of These' Trap

Similar Questions

Let the vectors $a, b, c$ represent three coterminous edges of a parallelopiped of volume $V$ . Then the volume of the parallelopiped, whose coterminous edges are represented by $a, b + c$ and $a + 2 b + 3 c$ is equal to

If the volume of a parallelopiped, whose coterminus edges are given by the vectors $a = \hat{i} + \hat{j} + n \hat{k}, b = 2 \hat{i} + 4 \hat{j} - n \hat{k}$ and $c = \hat{i} + n \hat{j} + 3 \hat{k} (n > 0)$ , is 158 cu.units, then :

The value of 'a' so that the volume of parallelopiped formed by $\hat{i} + a \hat{j} + \hat{k}$ , $\hat{j} + a \hat{k}$ and $a \hat{i} + \hat{k}$ becomes minimum is

If the volume of parallelopiped formed by the vectors $\hat{i} + λ \hat{j} + \hat{k}$ , $\hat{j} + λ \hat{k}$ and $λ \hat{i} + \hat{k}$ is minimum, then $λ$ is equal to :

If volume of parallelopiped whose coterminous edges are given by $u = \hat{i} + \hat{j} + λ \hat{k}$ , $v = \hat{i} + \hat{j} + 3 \hat{k}$ and $w = 2 \hat{i} + \hat{j} + \hat{k}$ be 1 cu. unit. If $θ$ be the angle between the edges $u$ and $w$ , then, $cos θ$ can be :

Match List I with List II:

List-I

List-II

Let the volume of a parallelepiped whose coterminous edges are given by $u = \hat{i} + \hat{j} + λ \hat{k}, v = \hat{i} + \hat{j} + 3 \hat{k}$ and $w = 2 \hat{i} + \hat{j} + \hat{k}$ be 1 cu. unit. If $θ$ be the angle between the edges $u$ and $w$ , then $cos θ$ can be :

If $a = α \hat{i} + β \hat{j} + 3 \hat{k}$ , $b = β \hat{i} - α \hat{j} - \hat{k}$ and $c = \hat{i} - 2 \hat{j} - \hat{k}$ such that $a \cdot b = 1$ and $b \cdot c = - 3$ , then $\frac{1}{3} ((a \times b) \cdot c)$ is equal to

Animated Solution for Mathematics - Vector Algebra: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The volume of the parallelopiped whose sides are given by OA=2i^−2j^​,OB=i^+j^​−k^,OC=3i^−k^, is

Select Answer:

Visualized Solution

Constructing the Parallelepiped

The Scalar Triple Product

Setting up the Determinant

Expanding along the First Row

Calculating the First Minor

Calculating the Second Minor

Combining the Terms

Comparing with Options

The Sigma Insight: Scalar Triple Product

Analyzing the Setup

The Master Tool

The Computational Engine

The 'None of These' Trap

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the volume of a parallelopiped, whose coterminus edges are given by the vectors a=i^+j^​+nk^,b=2i^+4j^​−nk^ and c=i^+nj^​+3k^(n>0), is 158 cu.units, then :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of 'a' so that the volume of parallelopiped formed by i^+aj^​+k^, j^​+ak^ and ai^+k^ becomes minimum is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the volume of parallelopiped formed by the vectors i^+λj^​+k^, j^​+λk^ and λi^+k^ is minimum, then λ is equal to :

Match List I with List II:

List-I

List-II

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a=αi^+βj^​+3k^, b=βi^−αj^​−k^ and c=i^−2j^​−k^ such that a⋅b=1 and b⋅c=−3, then 31​((a×b)⋅c) is equal to

Animated Solution for Mathematics - Vector Algebra: The volume of the parallelopiped whose sides are given by $O A = 2 \hat{i} - 2 \hat{j}, O B = \hat{i} + \hat{j} - \hat{k}, O C = 3 \hat{i} - \hat{k}$ , is

If the volume of a parallelopiped, whose coterminus edges are given by the vectors $a = \hat{i} + \hat{j} + n \hat{k}, b = 2 \hat{i} + 4 \hat{j} - n \hat{k}$ and $c = \hat{i} + n \hat{j} + 3 \hat{k} (n > 0)$ , is 158 cu.units, then :

The value of 'a' so that the volume of parallelopiped formed by $\hat{i} + a \hat{j} + \hat{k}$ , $\hat{j} + a \hat{k}$ and $a \hat{i} + \hat{k}$ becomes minimum is

If the volume of parallelopiped formed by the vectors $\hat{i} + λ \hat{j} + \hat{k}$ , $\hat{j} + λ \hat{k}$ and $λ \hat{i} + \hat{k}$ is minimum, then $λ$ is equal to :

If $a = α \hat{i} + β \hat{j} + 3 \hat{k}$ , $b = β \hat{i} - α \hat{j} - \hat{k}$ and $c = \hat{i} - 2 \hat{j} - \hat{k}$ such that $a \cdot b = 1$ and $b \cdot c = - 3$ , then $\frac{1}{3} ((a \times b) \cdot c)$ is equal to