1. **State the problem:** Given points $A(1,3)$ and $B(0,-1)$, construct the vector $\overrightarrow{AB}$ and find its magnitude and direction angle. 2. **Construct the vector $\overrightarrow{AB}$:** The vector from $A$ to $B$ is found by subtracting coordinates of $A$ from $B$: $$\overrightarrow{AB} = (x_B - x_A, y_B - y_A) = (0 - 1, -1 - 3) = (-1, -4)$$ 3. **Find the magnitude of $\overrightarrow{AB}$:** The magnitude (length) is given by the formula: $$|\overrightarrow{AB}| = \sqrt{(-1)^2 + (-4)^2} = \sqrt{1 + 16} = \sqrt{17}$$ 4. **Find the direction angle $\theta$:** The angle $\theta$ that $\overrightarrow{AB}$ makes with the positive x-axis is: $$\theta = \tan^{-1}\left(\frac{y}{x}\right) = \tan^{-1}\left(\frac{-4}{-1}\right) = \tan^{-1}(4)$$ Since both components are negative, the vector lies in the third quadrant, so add $180^\circ$ (or $\pi$ radians) to the angle: $$\theta = \tan^{-1}(4) + 180^\circ \approx 76^\circ + 180^\circ = 256^\circ$$ 5. **Final answers:** - Vector $\overrightarrow{AB} = (-1, -4)$ - Magnitude $|\overrightarrow{AB}| = \sqrt{17} \approx 4.123$ - Direction angle $\theta \approx 256^\circ$ (measured counterclockwise from positive x-axis)