A force of $$-F\hat{k}$$ acts on $$O$$, the origin of the coordinate system. The torque about the point $$(1, -1)$$ is

To find the torque about the given point, we can use the vector formula for torque:

$$\tau = \vec{r} \times \vec{F}$$

1. Identify the given vectors: The force acts at the origin $$O(0, 0, 0)$$, so the position vector of the point of application of force is:$$\vec{r}_O = 0\hat{i} + 0\hat{j} + 0\hat{k}$$

We need to find the torque about the point $$P(1, -1, 0)$$. The position vector of this point is:

$$\vec{r}_P = 1\hat{i} - 1\hat{j} + 0\hat{k}$$

The position vector $$\vec{r}$$ of the origin relative to point $$P$$ is:

$$\vec{r} = \vec{r}_O - \vec{r}_P$$

$$\vec{r} = (0 - 1)\hat{i} + (0 - (-1))\hat{j} + (0 - 0)\hat{k}$$

$$\vec{r} = -\hat{i} + \hat{j}$$

The given force vector is:

$$\vec{F} = -F\hat{k}$$

1. Calculate the cross product:$$\tau = (-\hat{i} + \hat{j}) \times (-F\hat{k})$$

Distributing the terms:

$$\tau = (-\hat{i} \times -F\hat{k}) + (\hat{j} \times -F\hat{k})$$

$$\tau = F(\hat{i} \times \hat{k}) - F(\hat{j} \times \hat{k})$$

Using the standard unit vector cross products ($$\hat{i} \times \hat{k} = -\hat{j}$$ and $$\hat{j} \times \hat{k} = \hat{i}$$):

$$\tau = F(-\hat{j}) - F(\hat{i})$$

$$\tau = -F\hat{i} - F\hat{j}$$

Factoring out $$-F$$:

$$\tau = -F(\hat{i} + \hat{j})$$