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The vector cross product is a fundamental operation in vector mathematics, often used in physics and engineering to find the torque exerted by a force. It involves two vectors \( \vec{A} \) and \( \vec{B} \), and the result is a new vector that is perpendicular to both of the original vectors. This is particularly valuable when calculating torque because it describes the rotational tendency of a force around an axis.

When calculating the vector cross product \( \vec{A} \times \vec{B} \), the magnitude of the product is given by \( |\vec{A}||\vec{B}|\sin(\theta) \), where \( \theta \) is the angle between the two vectors. The direction of the resultant vector can be determined using the right-hand rule. This product is expressed in terms of the unit vectors \( \hat{i}, \hat{j}, \hat{k} \), representing the x, y, and z axes respectively.

In practical problems, like the one presented here, setting up a cross product to find torque \( \vec{\tau} = \vec{r} \times \vec{F} \) involves working with the components of the vectors, often requiring the use of determinants to solve effectively. The resulting torque vector indicates how the force impacts the rotation of an object about an axis.

A determinant is a special number that can be calculated from the components of a square matrix. It gives useful information about the matrix itself and is particularly useful in calculating the cross product of vectors, like in our torque problem.

To find the cross product using a determinant, set up a 3x3 matrix with unit vectors \( \hat{i}, \hat{j}, \hat{k} \) in the first row, components of the position vector \( \vec{r} \) in the second row, and components of the force vector \( \vec{F} \) in the third row.

The structure of the determinant matrix looks like this:

• \[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ r_x & r_y & r_z \ F_x & F_y & F_z \end{vmatrix} \]

To actually compute the determinant, expand it using cofactor expansion along the first row. Each cofactor corresponds to a multiplication of a minor determinant and the unit vector. This yields a vector result in terms of these unit vectors.

Here, solving determinants allows us to clearly find the torque vector direction and magnitude resulting from the applied forces on the particle.

Solving physics problems often involves a careful approach that starts with understanding the physical context, breaking down the given data, applying relevant formulas, and performing accurate calculations. This problem, involving torque calculation, falls under classical mechanics where understanding the rotational effects of forces is crucial.

Begin with identifying the concepts involved. Here, torque is the primary concept, associated with rotational dynamics. A force applied to a body can cause it to rotate if the force isn't acting through the body's center of mass. The degree of rotation or torque depends on the magnitude of the force, the distance from the pivot (or axis), and the angle at which the force is applied.

Next, translate the physical scenario into mathematical expressions using the coordinate and force components provided. This includes setting up vectors for position and force, computing their cross product, and determining resultant vectors. Use known techniques such as the vector cross product and determinant calculation to find solutions.

Finally, interpret the results. In this case, understanding the computed torque vector helps to grasp the nature of particle rotation due to given forces, encapsulatiing concepts of directionality (given by vector components) and rotational magnitude. This approach not only aids in solving the given problem but also strengthens comprehension of physics at a fundamental level.