Let's analyze and solve the **first problem (1.59)** step by step using the image you provided. ## Step 1: **Write the vectors \(\vec{A}\), \(\vec{B}\), and \(\vec{C}\) in Cartesian coordinates** Let's the vectors from the graph in the image: - The **red vector \(\vec{A}\)** starts at the origin (0,0) and ends at (-2, 2). - The **blue vector \(\vec{B}\)** starts at the origin (0,0) and ends at (3, 1). - The **green vector \(\vec{C}\)** starts at the origin (0,0) and ends at (2, -3). ### **Cartesian Coordinates** - \(\vec{A} = \langle -2, 2 \rangle\) - \(\vec{B} = \langle 3, 1 \rangle\) - \(\vec{C} = \langle 2, -3 \rangle\) --- ## Step 2: **Calculate the length and direction of the vectors \(\vec{A}\), \(\vec{B}\), and \(\vec{C}\) (for 1.60)** Let's use the formula for the magnitude: \[ |\vec{V}| = \sqrt{V_x^2 + V_y^2} \] ### **Magnitude** - \(|\vec{A}| = \sqrt{(-2)^2 + (2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}\) - \(|\vec{B}| = \sqrt{(3)^2 + (1)^2} = \sqrt{9 + 1} = \sqrt{10}\) - \(|\vec{C}| = \sqrt{(2)^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13}\) ### **Direction (angle \(\theta\) from the x-axis, counterclockwise)** \[ \theta = \tan^{-1}\left(\frac{V_y}{V_x}\right) \] - \(\theta_A = \tan^{-1}\left(\frac{2}{-2}\right) = \tan^{-1}(-1) = 135^\circ\) (since it's in the second quadrant) - \(\theta_B = \tan^{-1}\left(\frac{1}{3}\right) \approx 18.4^\circ\) - \(\theta_C = \tan^{-1}\left(\frac{-3}{2}\right) \approx -56.3^\circ\) (or 303.7° if you prefer a positive angle) --- ## **Summary Table** | Vector | Cartesian Form | Magnitude | Direction (from x-axis) | |--------|----------------------|-----------------|------------------------| | \(\vec{A}\) | \(\langle -2, 2 \rangle\) | \(2\sqrt{2}\) | \(135^\circ\) | | \(\vec{B}\) | \(\langle 3, 1 \rangle\) | \(\sqrt{10}\) | \(18.4^\circ\) | | \(\vec{C}\) | \(\langle 2, -3 \rangle\) | \(\sqrt{13}\) | \(-56.3^\circ\) | --- Let me know if you want to proceed with the next steps (1.61, 1.62, and 1.63)!