Let's walk through the step-by-step solution for the given problem: ## Step 1: **Determine the Angles of the Vectors** Let's assume you have measured the angles (from the horizontal, counterclockwise) from the image. For demonstration, let's assume: - **Vector 1** (5.0 cm): \( \theta_1 = 127^\circ \) - **Vector 2** (3.5 cm): \( \theta_2 = 73^\circ \) - **Vector 3** (7.0 cm): \( \theta_3 = 10^\circ \) *(You should measure these angles yourself from the image for an exact answer!)* --- ## Step 2: **Write Each Vector in Component Form** ### General Formula: \[ \vec{A} = A_x \hat{i} + A_y \hat{j} = A \cos(\theta) \hat{i} + A \sin(\theta) \hat{j} \] ### Calculating Components: #### **Vector 1** (5.0 cm, \(127^\circ\)): \[ A_{1x} = 5.0 \cos(127^\circ) \] \[ A_{1y} = 5.0 \sin(127^\circ) \] #### **Vector 2** (3.5 cm, \(73^\circ\)): \[ A_{2x} = 3.5 \cos(73^\circ) \] \[ A_{2y} = 3.5 \sin(73^\circ) \] #### **Vector 3** (7.0 cm, \(10^\circ\)): \[ A_{3x} = 7.0 \cos(10^\circ) \] \[ A_{3y} = 7.0 \sin(10^\circ) \] --- ## Step 3: **Add the Vectors Using Components** Sum the x-components and y-components: \[ A_{x, \text{total}} = A_{1x} + A_{2x} + A_{3x} \] \[ A_{y, \text{total}} = A_{1y} + A_{2y} + A_{3y} \] **Calculate each:** - \( \cos(127^\circ) \approx -0.6018 \), \( \sin(127^\circ) \approx 0.7986 \) - \( \cos(73^\circ) \approx 0.2924 \), \( \sin(73^\circ) \approx 0.9563 \) - \( \cos(10^\circ) \approx 0.9848 \), \( \sin(10^\circ) \approx 0.1736 \) \[ A_{1x} = 5.0 \times -0.6018 = -3.009 \] \[ A_{1y} = 5.0 \times 0.7986 = 3.993 \] \[ A_{2x} = 3.5 \times 0.2924 = 1.023 \] \[ A_{2y} = 3.5 \times 0.9563 = 3.347 \] \[ A_{3x} = 7.0 \times 0.9848 = 6.894 \] \[ A_{3y} = 7.0 \times 0.1736 = 1.215 \] **Sum:** \[ A_{x,\text{total}} = -3.009 + 1.023 + 6.894 = 4.908 \] \[ A_{y,\text{total}} = 3.993 + 3.347 + 1.215 = 8.555 \] --- ## Step 4: **Resultant Magnitude and Direction** \[ |\vec{A}_{\text{total}}| = \sqrt{A_{x,\text{total}}^2 + A_{y,\text{total}}^2} = \sqrt{(4.908)^2 + (8.555)^2} = \sqrt{24.09 + 73.20} = \sqrt{97.29} \approx 9.86 \text{ cm} \] Direction (\(\theta\)) from x-axis: \[ \theta = \arctan \left(\frac{A_{y,\text{total}}}{A_{x,\text{total}}}\right) = \arctan\left(\frac{8.555}{4.908}\right) \approx \arctan(1.744) \approx 60.3^\circ \] --- ## Step 5: **Comparison of Methods** - If you add the vectors head-to-tail on paper, you should get the same result as above (within measurement error). --- ## Step 6: **What Vector to Add to Get Zero Vector?** The negative of the resultant: \[ \vec{A}_{\text{needed}} = -\vec{A}_{\text{total}} = -4.908 \hat{i} - 8.555 \hat{j} \] Magnitude: **9.86 cm**, Direction: **60.3° below the x-axis** (i.e., add \(180^\circ\) to the direction: \(60.3^\circ + 180^\circ = 240.3^\circ\)) --- ## **Final Answers** **(a) Resultant vector:** - Magnitude: **9.86 cm** - Direction: **60.3°** above the x-axis **(b) Vector needed to bring sum to zero:** - Magnitude: **9.86 cm** - Direction: **240.3°** (or **60.3° below the x-axis**) --- **Note:** - Use the actual measured angles from the image for your work! - The steps above show the complete method and how to use it with your measurements.

Certainly! Here's a detailed, step-by-step solution for each part of the problem, including explanations and final answers. --- ### **Part 1: Measure the Angles of the Vectors** **Instruction:** Using a protractor or smartphone app, measure the angles of the three vectors from the horizontal (x-axis). *(Since I cannot measure directly, I'll assume the angles based on typical measurements from the image. Please replace these with your actual measurements.)* Assumed measured angles: - **Vector 1:** \( \theta_1 = 127^\circ \) - **Vector 2:** \( \theta_2 = 73^\circ \) - **Vector 3:** \( \theta_3 = 10^\circ \) --- ### **Part 2: Write the Vectors in Component Form** Given magnitudes: - \( |\vec{A}_1| = 5.0\, \text{cm} \) - \( |\vec{A}_2| = 3.5\, \text{cm} \) - \( |\vec{A}_3| = 7.0\, \text{cm} \) **Component formulas:** \[ A_x = A \cos \theta, \quad A_y = A \sin \theta \] **Calculations:** 1. **Vector 1:** \[ A_{1x} = 5.0 \times \cos(127^\circ) \\ A_{1y} = 5.0 \times \sin(127^\circ) \] Using approximate trigonometric values: \[ \cos(127^\circ) \approx -0.6018 \\ \sin(127^\circ) \approx 0.7986 \] \[ A_{1x} = 5.0 \times -0.6018 = -3.009\, \text{cm} \\ A_{1y} = 5.0 \times 0.7986 = 3.993\, \text{cm} \] 2. **Vector 2:** \[ A_{2x} = 3.5 \times \cos(73^\circ) \\ A_{2y} = 3.5 \times \sin(73^\circ) \] \[ \cos(73^\circ) \approx 0.2924 \\ \sin(73^\circ) \approx 0.9563 \] \[ A_{2x} = 3.5 \times 0.2924 = 1.023\, \text{cm} \\ A_{2y} = 3.5 \times 0.9563 = 3.347\, \text{cm} \] 3. **Vector 3:** \[ A_{3x} = 7.0 \times \cos(10^\circ) \\ A_{3y} = 7.0 \times \sin(10^\circ) \] \[ \cos(10^\circ) \approx 0.9848 \\ \sin(10^\circ) \approx 0.1736 \] \[ A_{3x} = 7.0 \times 0.9848 = 6.894\, \text{cm} \\ A_{3y} = 7.0 \times 0.1736 = 1.215\, \text{cm} \] --- ### **Part 3: Add the Vectors Using Two Methods** #### **Method 1: Head-to-Tail Method** - Draw vector 1, then from its tip draw vector 2, then from the tip of vector 2 draw vector 3. - The resultant vector is from the tail of vector 1 to the tip of vector 3. - Measure the magnitude and direction of this resultant on paper. *(Since I cannot draw, I will proceed with component addition which yields the same result.)* --- #### **Method 2: Add Components** Sum x-components: \[ A_{x,\text{total}} = A_{1x} + A_{2x} + A_{3x} = -3.009 + 1.023 + 6.894 = 4.908\, \text{cm} \] Sum y-components: \[ A_{y,\text{total}} = A_{1y} + A_{2y} + A_{3y} = 3.993 + 3.347 + 1.215 = 8.555\, \text{cm} \] **Magnitude of the resultant vector:** \[ |\vec{A}_{\text{total}}| = \sqrt{(4.908)^2 + (8.555)^2} \] \[ = \sqrt{24.09 + 73.20} = \sqrt{97.29} \approx 9.86\, \text{cm} \] **Direction (angle with respect to x-axis):** \[ \theta = \arctan \left( \frac{A_{y,\text{total}}}{A_{x,\text{total}}} \right) = \arctan \left( \frac{8.555}{4.908} \right) \] \[ \approx \arctan(1.744) \approx 60.3^\circ \] --- ### **Part 4: Compare the Results** **Are the two results equal?** - Yes. The vector addition via the head-to-tail method and component addition should give the same magnitude and direction within measurement errors. --- ### **Part 5: Find the Vector to Add for Zero Result** To make the total vector zero, we need to add a vector that is equal in magnitude but opposite in direction to the resultant: \[ \vec{A}_{\text{needed}} = - \vec{A}_{\text{total}} \] Components: \[ A_{x,\text{needed}} = -4.908 \, \text{cm} \] \[ A_{y,\text{needed}} = -8.555 \, \text{cm} \] **Magnitude of this vector:** \[ |\vec{A}_{\text{needed}}| = \sqrt{(-4.908)^2 + (-8.555)^2} = 9.86\, \text{cm} \] **Direction (angle from x-axis):** \[ \theta_{\text{needed}} = \arctan \left( \frac{-8.555}{-4.908} \right) = \arctan(1.744) \] Since both components are negative, the vector points in the third quadrant: \[ \theta_{\text{needed}} = 180^\circ + 60.3^\circ = 240.3^\circ \] *(Alternatively, measure the angle as 60.3° below the negative x-axis.)* --- ### **Final Answers:** - **Resultant Vector (sum of three vectors):** - **Magnitude:** **9.86 cm** - **Direction:** **60.3° above the x-axis** - **Vector to add to make total zero:** - **Magnitude:** **9.86 cm** - **Direction:** **240.3° from the x-axis** (or 60.3° below the negative x-axis) --- **Note:** Ensure to measure the actual angles from the image for precise calculations. The above is a comprehensive example demonstrating the process and how to interpret the results.