Let \(\mathbf{F}=2 \mathbf{i}-3 \mathbf{j}+\mathbf{k}\) act at the point (5,1,3) (a) Find the torque of \(\mathbf{F}\) about the point (4,1,0) (b) Find the torque of \(\mathbf{F}\) about the line \(\mathbf{r}=4 \mathbf{i}+\mathbf{j}+(2 \mathbf{i}+\mathbf{j}-2 \mathbf{k}) t\).

The vector cross product is a mathematical operation that takes two vectors and returns a third vector that is perpendicular to the plane of the first two. This operation is essential in physics for calculating quantities like torque. For vectors \(\textbf{A}\) and \(\textbf{B}\), the cross product \(\textbf{A} \times \textbf{B}\) is computed using the determinant of a matrix. The formula looks like this: \[\(\textbf{A} \times \textbf{B}\) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ A_x & A_y & A_z \ B_x & B_y & B_z \end{vmatrix}\].
Here, \(\textbf{i}, \textbf{j},\) and \(\textbf{k}\) are unit vectors in the x, y, and z directions, respectively. The result is another vector whose x, y, and z components can be determined from the determinant.

A position vector represents the position of a point in space relative to an origin. In the exercise, you need to find the position vector \(\textbf{r}\) from one point (4,1,0) to another point (5,1,3). This is done by subtracting the coordinates component-wise: \[ \textbf{r} = (5 - 4) \textbf{i} + (1 - 1) \textbf{j} + (3 - 0) \textbf{k} = \textbf{i} + 0 \textbf{j} + 3 \textbf{k}\]
The resulting vector \(\textbf{r}\) indicates not only the magnitude but also the direction from the origin to the point.

The projection of one vector onto another vector is a critical step when computing the torque about a line. To project vector \(\textbf{r}\) onto the direction vector \(\textbf{d}\), use the formula: \[\textbf{r}_\text{proj} = \frac{\textbf{r} \cdot \textbf{d}}{\textbf{d} \cdot \textbf{d}} \textbf{d} \]
This formula finds how much of \(\textbf{r}\) lies along \(\textbf{d}\). The perpendicular component is then \(\textbf{r}_\text{perp} = \textbf{r} - \textbf{r}_\text{proj}\). This component is critical for calculating torque about a line.

A direction vector provides the orientation of a line or path in space. In this exercise, the direction vector \(\textbf{d}\) is derived from the line equation: \[ \textbf{r} = 4 \textbf{i} + \textbf{j} + (2 \textbf{i} + \textbf{j} - 2 \textbf{k}) t \]
The coefficient of t, \((2 \textbf{i} + \textbf{j} - 2 \textbf{k})\), is the direction vector. This vector determines the direction and sense of the line. It is important when projecting the position vector and calculating torque.