The image you uploaded presents a vector-related problem where you're asked to find a unit vector in the same direction as vector $\mathbf{v}$. The vector $\mathbf{v}$ has the components given by the coordinates of two points, $P(2, 3)$ and $Q(-2, -6)$, as shown in the diagram.

Steps to Solve:

1. Find the Vector $\mathbf{v}$: The vector $\mathbf{v}$ can be determined by subtracting the coordinates of point $P$ from $Q$: $\mathbf{v} = Q - P = (-2 - 2, -6 - 3) = (-4, -9)$

2. Calculate the Magnitude of $\mathbf{v}$: The magnitude (or length) of vector $\mathbf{v}$ is given by: $|\mathbf{v}| = \sqrt{(-4)^2 + (-9)^2} = \sqrt{16 + 81} = \sqrt{97}$

3. Find the Unit Vector in the Direction of $\mathbf{v}$: A unit vector in the same direction as $\mathbf{v}$ is obtained by dividing $\mathbf{v}$ by its magnitude: $\mathbf{u} = \frac{\mathbf{v}}{|\mathbf{v}|} = \left(\frac{-4}{\sqrt{97}}, \frac{-9}{\sqrt{97}}\right)$

Final Answer:

The unit vector in the direction of $\mathbf{v}$ is: $\mathbf{u} = \left(\frac{-4}{\sqrt{97}}, \frac{-9}{\sqrt{97}}\right)$

Would you like further details or have any questions?

Here are five related questions to expand your understanding:

1. How would the unit vector change if the coordinates of points $P$ and $Q$ were different?
2. What is the significance of a unit vector in physics or engineering?
3. How would you find a unit vector if given a vector in three-dimensional space?
4. What role does the magnitude of a vector play in normalizing it to a unit vector?
5. Can a unit vector have negative components? Why or why not?

Tip: Always double-check the sign and magnitude calculations when dealing with vectors to avoid errors in the final unit vector.