91Ó°ÊÓ

To compute the cross product of \(\mathbf{j}\) and \(\mathbf{k}\), we can use the following formula: $$\mathbf{j} \times \mathbf{k} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{vmatrix}$$ Here, \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\) are the standard basis vectors, and the determinant of the matrix represents the coefficients of each basis vector in the cross product result.

We can now compute the determinant to find the coefficients of the cross product: $$\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{vmatrix} = \mathbf{i} \begin{vmatrix} 1 & 0 \\ 0 & 1 \\ \end{vmatrix} - \mathbf{j} \begin{vmatrix} 0 & 0 \\ 0 & 1 \\ \end{vmatrix} + \mathbf{k} \begin{vmatrix} 0 & 1 \\ 0 & 0 \\ \end{vmatrix}$$

Compute the product of each basis vector with its respective sub-determinant: $$\mathbf{i} (1 - 0) - \mathbf{j} (0) + \mathbf{k} (0) = \mathbf{i}$$ So the cross product of \(\mathbf{j}\) and \(\mathbf{k}\) is \(\mathbf{i}\).

Now that we have computed the cross product, we can sketch the given vectors, \(\mathbf{j}\), and \(\mathbf{k}\), along with their cross product, \(\mathbf{i}\). 1. Begin by drawing a coordinate system with x, y, and z axis. This will represent the Cartesian coordinate system. 2. Now draw the \(\mathbf{j}\) vector, starting at the origin, extending in the positive y direction by one unit. 3. Next, draw the \(\mathbf{k}\) vector, starting at the origin, extending in the positive z direction by one unit. 4. Lastly, draw the cross product \(\mathbf{i}\), starting at the origin, extending in the positive x direction by one unit. Make sure that the angle between each pair of vectors is 90 degrees, which is indicative of them being orthogonal to each other.