Find the magnitude of the vector sum [tex]\overrightarrow{A} + \overrightarrow{B}[/tex].

Vector [tex]\overrightarrow{A}[/tex] is 17.4 m long in a 303° direction. Vector [tex]\overrightarrow{B}[/tex] is 25.6 m long in a 22.0° direction.



Answer :

To find the magnitude of the vector sum [tex]\(\overrightarrow{A} + \overrightarrow{B}\)[/tex], we must first break down each vector into its x and y components, add those components together, and finally determine the magnitude of the resultant vector.

1. Vector A Components:
- Magnitude of Vector A ([tex]\(|\overrightarrow{A}|\)[/tex]): [tex]\(17.4 \, \text{m}\)[/tex]
- Direction of Vector A: [tex]\(303^{\circ}\)[/tex]

The angle needs to be converted to radians for component calculations. However, we can directly use the given angles.

- The x-component of Vector A ([tex]\(A_x\)[/tex]) is found by:
[tex]\[ A_x = 17.4 \cdot \cos(303^{\circ}) \approx 9.477 \, \text{m} \][/tex]
- The y-component of Vector A ([tex]\(A_y\)[/tex]) is found by:
[tex]\[ A_y = 17.4 \cdot \sin(303^{\circ}) \approx -14.593 \, \text{m} \][/tex]

2. Vector B Components:
- Magnitude of Vector B ([tex]\(|\overrightarrow{B}|\)[/tex]): [tex]\(25.6 \, \text{m}\)[/tex]
- Direction of Vector B: [tex]\(22.0^{\circ}\)[/tex]

As with Vector A, we can directly use the given angles.

- The x-component of Vector B ([tex]\(B_x\)[/tex]) is found by:
[tex]\[ B_x = 25.6 \cdot \cos(22.0^{\circ}) \approx 23.736 \, \text{m} \][/tex]
- The y-component of Vector B ([tex]\(B_y\)[/tex]) is found by:
[tex]\[ B_y = 25.6 \cdot \sin(22.0^{\circ}) \approx 9.590 \, \text{m} \][/tex]

3. Components of the Resultant Vector [tex]\(\overrightarrow{R} = \overrightarrow{A} + \overrightarrow{B}\)[/tex]:
- The x-component of the resultant vector ([tex]\(R_x\)[/tex]) is:
[tex]\[ R_x = A_x + B_x \approx 9.477 + 23.736 = 33.213 \, \text{m} \][/tex]
- The y-component of the resultant vector ([tex]\(R_y\)[/tex]) is:
[tex]\[ R_y = A_y + B_y \approx -14.593 + 9.590 = -5.003 \, \text{m} \][/tex]

4. Magnitude of the Resultant Vector:
- The magnitude of the resultant vector ([tex]\(|\overrightarrow{R}|\)[/tex]) is found using the Pythagorean theorem:
[tex]\[ |\overrightarrow{R}| = \sqrt{R_x^2 + R_y^2} \approx \sqrt{(33.213)^2 + (-5.003)^2} \approx 33.587 \, \text{m} \][/tex]

Thus, the magnitude of the vector sum [tex]\(\overrightarrow{A} + \overrightarrow{B}\)[/tex] is approximately [tex]\(33.587 \, \text{m}\)[/tex].