91Ó°ÊÓ

To find the cross product of two vectors, we can use the formula: $$\mathbf{a} \times \mathbf{b} = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix} \times \begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix} = \begin{pmatrix} a_2 b_3 - a_3 b_2 \\ a_3 b_1 - a_1 b_3 \\ a_1 b_2 - a_2 b_1 \end{pmatrix}$$ In this case, we have \(\mathbf{a} = \mathbf{i} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\) and \(\mathbf{b} = \mathbf{k} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\). Applying the formula, we get: $$\mathbf{i} \times \mathbf{k} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \times \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ -1 \\ 0 \end{pmatrix} = -\mathbf{j}$$ So, the cross product of i and k is -j.

In a 3D coordinate system, the unit vectors i, j, and k represent the x, y, and z axes, respectively. To visualize the cross product, we will draw i, k and their cross product -j. 1. Draw the x, y, and z axes. 2. Draw the vector i, which goes along the x-axis (positive x direction). 3. Draw the vector k, which goes along the z-axis (positive z direction). 4. Draw the cross product -j as a vector going in the opposite direction of the y-axis (negative y direction). In this sketch, we can observe that the cross product of i and k is perpendicular to both vectors, which is one of the key properties of cross products. The magnitude of the resulting cross product is equal to one, as both i and k are unit vectors.