Problem 49 - Master Resource Book in JEE Main Physics

Question

The torque of a force \(\mathbf{F}=-3 \hat{\mathbf{i}}+\hat{\mathbf{j}}+5 \hat{\mathbf{k}}\) acting at a point is \(\tau .\) If the position vector of the point is \(7 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}}\), then \(\tau\) is (a) \(7 \hat{\mathrm{i}}-8 \hat{\mathrm{j}}+9 \hat{\mathrm{k}}\) (b) \(14 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}}\) (c) \(2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+8 \hat{\mathrm{k}}\) (d) \(14 \hat{\mathrm{i}}-38 \hat{\mathrm{j}}+16 \hat{\mathrm{k}}\)

Step-by-Step Solution

Answer

The torque \( \boldsymbol{\tau} \) is option (d) \( 14 \hat{\mathrm{i}} - 38 \hat{\mathrm{j}} + 16 \hat{\mathrm{k}} \).

Step 1: Understand the Torque Formula

The torque \( \boldsymbol{\tau} \) is given by the cross product of the position vector \( \mathbf{r} \) and the force vector \( \mathbf{F} \), i.e., \( \boldsymbol{\tau} = \mathbf{r} \times \mathbf{F} \). We need to compute this cross product to find \( \boldsymbol{\tau} \).

Step 2: Set Up the Cross Product

Write down the given vectors. The position vector is \( \mathbf{r} = 7 \hat{\mathbf{i}} + 3 \hat{\mathbf{j}} + \hat{\mathbf{k}} \) and the force vector is \( \mathbf{F} = -3 \hat{\mathbf{i}} + \hat{\mathbf{j}} + 5 \hat{\mathbf{k}} \).

Step 3: Calculate the Determinant

To find \( \mathbf{r} \times \mathbf{F} \), compute the determinant:\[\begin{vmatrix}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \7 & 3 & 1 \-3 & 1 & 5 \\end{vmatrix}\]Use cofactor expansion along the first row to solve the determinant.

Step 4: Expand the Determinant

Compute the determinant with cofactor expansion:\[ \boldsymbol{\tau} = \hat{\mathbf{i}}(3 \cdot 5 - 1 \cdot 1) - \hat{\mathbf{j}}(7 \cdot 5 - 1 \cdot (-3)) + \hat{\mathbf{k}}(7 \cdot 1 - 3 \cdot (-3)) \]

Step 5: Simplify the Calculations

Perform the multiplications and subtractions:\[ = \hat{\mathbf{i}}(15 - 1) - \hat{\mathbf{j}}(35 + 3) + \hat{\mathbf{k}}(7 + 9) \]\[ = 14\hat{\mathbf{i}} - 38\hat{\mathbf{j}} + 16\hat{\mathbf{k}} \]

Step 6: Choose the Correct Option

Compare the calculated torque \( \boldsymbol{\tau} = 14\hat{\mathbf{i}} - 38\hat{\mathbf{j}} + 16\hat{\mathbf{k}} \) with the given options. The correct answer is (d) \( 14 \hat{\mathrm{i}} - 38 \hat{\mathrm{j}} + 16 \hat{\mathrm{k}} \).

Key Concepts

Cross Product

In the realm of physics and vector calculus, the cross product is a crucial operation used to determine the relationship between two vectors in three-dimensional space. In the context of calculating torque, the cross product provides a vector that represents the axis of rotation and magnitude of the torque. For vectors \( \mathbf{A} = a_1 \hat{\mathbf{i}} + a_2 \hat{\mathbf{j}} + a_3 \hat{\mathbf{k}} \) and \( \mathbf{B} = b_1 \hat{\mathbf{i}} + b_2 \hat{\mathbf{j}} + b_3 \hat{\mathbf{k}} \), their cross product \( \mathbf{A} \times \mathbf{B} \) is defined as:

• \( c_1 = a_2b_3 - a_3b_2 \)
• \( c_2 = a_3b_1 - a_1b_3 \)
• \( c_3 = a_1b_2 - a_2b_1 \)

The resulting vector, \( \mathbf{C} = c_1 \hat{\mathbf{i}} + c_2 \hat{\mathbf{j}} + c_3 \hat{\mathbf{k}} \), is perpendicular to both \( \mathbf{A} \) and \( \mathbf{B} \). This property is especially useful for insights like torque, where the rotational effect of a force about a pivot is considered. It helps create a clear depiction of the direction and magnitude of the reaction without ambiguity in three-dimensional space.
Understanding the cross product requires grasping the geometric intuition behind it: it's all about finding a vector that not only charts the plane containing your initial two vectors but also points perpendicularly out of that plane.

Vector Calculus

Vector calculus is an essential component of understanding physical phenomena expressed through vectors, such as force, displacement, and velocity. It extends calculus from scalar fields to vector fields, bringing in new operations like differentiation and integration of vectors, as well as vector products like the dot product and the cross product.
In the context of the exercise, vector calculus enables us to find the torque (\( \boldsymbol{\tau} \)), a vector quantity, from the cross product of the force vector \( \mathbf{F} \) and the position vector \( \mathbf{r} \). This is crucial in physics since torque quantifies rotational force, essential in understanding how objects move and interact under the influence of forces in a rotational context.
Vector calculus also supports a variety of applications beyond just torque. It includes paths, surfaces, flux through a surface, and potential energies. These distinctions enable professionals to project real-world physical systems within the structures of mathematics, solving complex problems and making predictions.

• Gradient: Indicates the rate and direction of change in a scalar field.
• Divergence: Provides a measure of a vector field's source or sink at a given point.
• Curl: Represents a vector field's rotation around a point.

Mastering these operations is essential for delving deeper into physics topics like electromagnetism and fluid dynamics, where fields vary over space.

Determinant Calculation

Calculated within the operations of linear algebra, determinants are vital for assessing vectors' properties in multidimensional spaces. In the exercise, the determinant is part of determining the cross product between the position vector and the force vector, which leads to developing the torque vector.
The determinant of a 3x3 matrix is essential in calculating cross products:\[\begin{vmatrix}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \7 & 3 & 1 \-3 & 1 & 5 \\end{vmatrix}\]Expanding this determinant with cofactor expansion along the first row results in a methodical subtraction and multiplication, contributing to the components of the resulting vector:

• For \( \hat{\mathbf{i}} \), perform: \ \(3 \cdot 5 - 1 \cdot 1 = 14 \).
• For \( \hat{\mathbf{j}} \), perform: \ \(7 \cdot 5 - 1 \cdot (-3) = -38 \).
• For \( \hat{\mathbf{k}} \), perform: \ \(7 \cdot 1 - 3 \cdot (-3) = 16 \).

The determinant represents a "volume" encapsulated by the vectors acting as edges, providing not only solutions through calculation but also insights into geometry and orientation within the space.
Practicing determinant calculations equips learners with tools to solve systems of equations, understand vector spaces, and explore properties like linear independence.