The image appears to contain a question related to vectors involving a triangle, but it's somewhat rotated. I'll rotate and analyze the image to provide the correct solution.

Analyzing the Problem:

Given Data:

• Triangle PQR is provided with vectors $\overrightarrow{PQ} = \mathbf{a}$ and $\overrightarrow{PR} = \mathbf{b}$.
• A point S lies on QR such that $QS : SR = 1 : 2$.
• You're required to:
  1. Express $\overrightarrow{QS}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.
  2. Find $\overrightarrow{PS}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.

Step-by-Step Solution:

1. Expressing $\overrightarrow{QS}$:

Given $QS : SR = 1 : 2$, $S$ divides $QR$ in the ratio $1:2$.

Let $\overrightarrow{QR} = \mathbf{c}$.

Now, $\overrightarrow{QS}$ can be found using the section formula: $\overrightarrow{QS} = \frac{2\overrightarrow{Q} + 1\overrightarrow{R}}{1+2} = \frac{2\overrightarrow{Q} + \overrightarrow{R}}{3}$

But $\overrightarrow{QR} = \overrightarrow{PR} - \overrightarrow{PQ} = \mathbf{b} - \mathbf{a}$.

So, $\overrightarrow{QS}$ becomes: $\overrightarrow{QS} = \frac{2(\overrightarrow{PR} - \overrightarrow{PQ}) + \overrightarrow{PR}}{3}$

Substitute $\mathbf{b}$ and $\mathbf{a}$: $\overrightarrow{QS} = \frac{2(\mathbf{b} - \mathbf{a}) + \mathbf{b}}{3} = \frac{2\mathbf{b} - 2\mathbf{a} + \mathbf{b}}{3} = \frac{3\mathbf{b} - 2\mathbf{a}}{3}$ $\overrightarrow{QS} = \mathbf{b} - \frac{2}{3}\mathbf{a}$

2. Expressing $\overrightarrow{PS}$:

$\overrightarrow{PS}$ is given by: $\overrightarrow{PS} = \overrightarrow{PQ} + \overrightarrow{QS}$

Substitute the expression for $\overrightarrow{QS}$: $\overrightarrow{PS} = \mathbf{a} + \left(\mathbf{b} - \frac{2}{3}\mathbf{a}\right)$ $\overrightarrow{PS} = \mathbf{b} + \left(\mathbf{a} - \frac{2}{3}\mathbf{a}\right) = \mathbf{b} + \frac{1}{3}\mathbf{a}$

Final Answers:

1. $\overrightarrow{QS} = \mathbf{b} - \frac{2}{3}\mathbf{a}$
2. (\overrightarrow{PS} = \frac{1}{3}\mathbf{a} + \mathbf{b