Problem 87

To solve the problem, we perform the following steps:

1. **Determine the vectors $\overrightarrow{AB}$ and $\overrightarrow{BC}$**

Given $A(1,1)$, $B(0,2)$, $C(-1,-1)$, we calculate:

- $\overrightarrow{AB} = B - A = (0-1, 2-1) = (-1, 1)$
- $\overrightarrow{BC} = C - B = (-1-0, -1-2) = (-1, -3)$

2. **Calculate the magnitudes of $\overrightarrow{AB}$ and $\overrightarrow{BC}$**

- $|\overrightarrow{AB}| = \sqrt{(-1)^2 + 1^2} = \sqrt{2}$
- $|\overrightarrow{BC}| = \sqrt{(-1)^2 + (-3)^2} = \sqrt{10}$

3. **Find the cosine of the angle between $\overrightarrow{AB}$ and $\overrightarrow{BC}$**

The cosine of the angle can be calculated by the formula:
$$\cos(\overrightarrow{AB},\overrightarrow{BC}) = \frac{\overrightarrow{AB} \cdot \overrightarrow{BC}}{|\overrightarrow{AB}| \cdot |\overrightarrow{BC}|}$$

Substituting the values, we get:
$$\cos(\overrightarrow{AB},\overrightarrow{BC}) = \frac{(-1)(-1) + (1)(-3)}{\sqrt{2} \cdot \sqrt{10}} = \frac{1 - 3}{\sqrt{2} \cdot \sqrt{10}} = -\frac{\sqrt{5}}{5}$$

This negative value indicates that the angle between $\overrightarrow{AB}$ and $\overrightarrow{BC}$ is obtuse, meaning the projection vector will be in the opposite direction of $\overrightarrow{BC}$.

4. **Calculate the magnitude of the projection of $\overrightarrow{AB}$ on $\overrightarrow{BC}$**

The magnitude of this projection can be found by:
$$|{\overrightarrow{AB}}| \cdot |\cos(\overrightarrow{AB},\overrightarrow{BC})| = \sqrt{2} \cdot \frac{\sqrt{5}}{5} = \frac{\sqrt{10}}{5}$$

Since the projection is in the opposite direction of $\overrightarrow{BC}$, we find the unit vector of $\overrightarrow{BC}$ and scale it by the magnitude of the projection:

- First, find the scale factor: $\frac{\frac{\sqrt{10}}{5}}{\sqrt{10}} = \frac{1}{5}$
- Then, apply the scale factor to $\overrightarrow{BC}$: $-\frac{1}{5} \cdot (-1, -3) = (\frac{1}{5}, \frac{3}{5})$

Therefore, the projection vector of $\overrightarrow{AB}$ on $\overrightarrow{BC}$ is $\boxed{(\frac{1}{5}, \frac{3}{5})}$.

Hence, the correct answer is $\boxed{\text{C}}$.