A force of $-\mathrm{P} \hat{\mathrm{k}}$ acts on the origin of the coordinate system. The torque about the point $(2,-3)$ is $\mathrm{P}(a \hat{i}+b \hat{j})$, The ratio of $\frac{a}{b}$ is $\frac{x}{2}$. The value of $x$ is -

<p>Let the point where the force acts be A, the origin of the coordinate system (0, 0, 0), and let the point about which the torque is calculated be B (2, -3, 0). The force vector is given by $\vec{F} = -P\hat{k}$.</p> <p>To find the torque, we first find the position vector of point A with respect to point B:</p> <p>$$\vec{r}_{AB} = \vec{r}_A - \vec{r}_B = (0 - 2)\hat{i} + (0 - (-3))\hat{j} + (0 - 0)\hat{k} = -2\hat{i} + 3\hat{j}$$</p> To calculate the cross product, we can use the determinant method with a 3x3 matrix: <br/><br/> $$\vec{\tau} = \vec{r}_{AB} \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 3 & 0 \\ 0 & 0 & -P \\ \end{vmatrix}$$ <br/><br/> Now, we will calculate the cross product components by expanding the determinant along the first row: <br/><br/> 1. $$\tau_i = \hat{i} \begin{vmatrix} 3 & 0 \\ 0 & -P \\ \end{vmatrix} = \hat{i}((3)(-P) - (0)(0)) = -3P\hat{i}$$ <br/><br/> 2. $$\tau_j = -\hat{j} \begin{vmatrix} -2 & 0 \\ 0 & -P \\ \end{vmatrix} = -\hat{j}((-2)(-P) - (0)(0)) = -2P\hat{j}$$<br/><br/> (Notice the negative sign in front of the $\hat{j}$ term, as it comes from the expansion of the determinant.) <br/><br/> 3. $$\tau_k = \hat{k} \begin{vmatrix} -2 & 3 \\ 0 & 0 \\ \end{vmatrix} = \hat{k}((-2)(0) - (3)(0)) = 0\hat{k}$$<br/><br/> Now, combine the components to get the torque vector: <br/><br/> $\vec{\tau} = -3P\hat{i} - 2P\hat{j} + 0\hat{k} = -3P\hat{i} - 2P\hat{j}$<br/> <p>Comparing this to the given torque vector $\vec{\tau} = P(a\hat{i} + b\hat{j})$, we find that:</p> <p>$a = -3$<br/><br/> $b = -2$</p> <p>Thus, the ratio $\frac{a}{b} = \frac{-3}{-2} = \frac{x}{2}$.<br/><br/> Therefore, $x = 3$.</p>