Let \( \mathbf{r}=\overrightarrow{\mathrm{OP}}=\mathbf{i}+\mathbf{j}-\mathbf{k} \). A force \( F=\langle 0,19,0\rangle \) is applied at \( P \). Find the torque \( \tau \) about \( O \) that is produced. \( \tau=\langle\square . \square \) (Simplify your answers. Type integers or fractions.)

To find the torque \( \tau \) about point \( O \) produced by the force \( F \) applied at point \( P \), we can use the formula for torque: \[ \tau = \mathbf{r} \times \mathbf{F} \] where \( \mathbf{r} = \langle 1, 1, -1 \rangle \) and \( \mathbf{F} = \langle 0, 19, 0 \rangle \). We can calculate the cross product \( \mathbf{r} \times \mathbf{F} \) using the determinant method: \[ \tau = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & -1 \\ 0 & 19 & 0 \end{vmatrix} \] Calculating this determinant, we have: \[ \tau = \mathbf{i} \begin{vmatrix} 1 & -1 \\ 19 & 0 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 1 & -1 \\ 0 & 0 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 1 & 1 \\ 0 & 19 \end{vmatrix} \] Now, we evaluate each of the 2x2 determinants: 1. For \( \mathbf{i} \): \[ \begin{vmatrix} 1 & -1 \\ 19 & 0 \end{vmatrix} = (1)(0) - (-1)(19) = 0 + 19 = 19 \] 2. For \( \mathbf{j} \): \[ \begin{vmatrix} 1 & -1 \\ 0 & 0 \end{vmatrix} = (1)(0) - (-1)(0) = 0 \] 3. For \( \mathbf{k} \): \[ \begin{vmatrix} 1 & 1 \\ 0 & 19 \end{vmatrix} = (1)(19) - (1)(0) = 19 \] Putting it all together: \[ \tau = 19\mathbf{i} - 0\mathbf{j} + 19\mathbf{k} = \langle 19, 0, 19 \rangle \] Thus, the torque \( \tau \) about point \( O \) is given by: \[ \tau = \langle 19, 0, 19 \rangle \]