91Ó°ÊÓ

Compute the cross product of -\(\mathbf{j}\) and \(\mathbf{k}\) using the standard unit vectors, which are represented by \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\). In general, the cross product between two vectors \(\mathbf{A}\) and \(\mathbf{B}\) is given by the determinant of a matrix formed using the standard unit vectors and the components of both vectors: $$\mathbf{A} \times \mathbf{B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix}$$ Applying this formula to the given problem of finding the cross product of -\(\mathbf{j}\) and \(\mathbf{k}\), where -\(\mathbf{j}\) has components (0, -1, 0) and \(\mathbf{k}\) has components (0, 0, 1): $$(-\mathbf{j}) \times \mathbf{k} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{vmatrix}$$ Calculating the determinant, we get: $$(-\mathbf{j}) \times \mathbf{k} = \mathbf{i}((-1)(1)) - \mathbf{j}(0) + \mathbf{k}(0) = \mathbf{i}$$ So, the cross product of -\(\mathbf{j}\) and \(\mathbf{k}\) is \(\mathbf{i}\).

Next, we will sketch the two given vectors -\(\mathbf{j}\) and \(\mathbf{k}\), as well as their cross product \(\mathbf{i}\). 1. First, we will draw the coordinate axes (x, y, and z) to represent our 3D space. 2. Then, we will draw the vector -\(\mathbf{j}\) as a line segment going down from the origin, representing its direction and magnitude in the negative y-axis direction. 3. Next, we will draw the vector \(\mathbf{k}\) as a line segment extending from the origin in the positive z-axis direction. 4. Finally, we will draw the cross product of these vectors, which is \(\mathbf{i}\), as a line segment extending from the origin in the positive x-axis direction. The sketch should clearly show that \(\mathbf{i}\) is perpendicular to both -\(\mathbf{j}\) and \(\mathbf{k}\).