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

Vector [tex]\(\vec{A}\)[/tex] is 448 m long in a [tex]\(224^{\circ}\)[/tex] direction.
Vector [tex]\(\vec{B}\)[/tex] is 336 m long in a [tex]\(75.9^{\circ}\)[/tex] direction.



Answer :

To find the magnitude of the vector sum [tex]\(\vec{A} + \vec{B}\)[/tex], we need to proceed step-by-step by breaking down the vectors into their components, summing the components, and then finding the resultant vector's magnitude.

### Step 1: Break down each vector into its components

1. Vector [tex]\(\vec{A}\)[/tex]
- Magnitude: [tex]\(A = 448\)[/tex] m
- Direction: [tex]\(224^{\circ}\)[/tex]

To find the x and y components of [tex]\(\vec{A}\)[/tex]:
[tex]\[ A_x = A \cos(224^{\circ}) \approx -322.26 \text{ m} \][/tex]
[tex]\[ A_y = A \sin(224^{\circ}) \approx -311.21 \text{ m} \][/tex]

2. Vector [tex]\(\vec{B}\)[/tex]
- Magnitude: [tex]\(B = 336\)[/tex] m
- Direction: [tex]\(75.9^{\circ}\)[/tex]

To find the x and y components of [tex]\(\vec{B}\)[/tex]:
[tex]\[ B_x = B \cos(75.9^{\circ}) \approx 81.85 \text{ m} \][/tex]
[tex]\[ B_y = B \sin(75.9^{\circ}) \approx 325.88 \text{ m} \][/tex]

### Step 2: Sum the components of the vectors

To find the components of the resultant vector [tex]\(\vec{R} = \vec{A} + \vec{B}\)[/tex] :

1. x-component:
[tex]\[ R_x = A_x + B_x = -322.26 + 81.85 \approx -240.41 \text{ m} \][/tex]

2. y-component:
[tex]\[ R_y = A_y + B_y = -311.21 + 325.88 \approx 14.67 \text{ m} \][/tex]

### Step 3: Find the magnitude of the resultant vector

The magnitude of the resultant vector [tex]\(\vec{R}\)[/tex] can be found using the Pythagorean theorem:
[tex]\[ R = \sqrt{R_x^2 + R_y^2} \][/tex]

Plugging in the values we have:
[tex]\[ R = \sqrt{(-240.41)^2 + (14.67)^2} \approx 240.86 \text{ m} \][/tex]

So, the magnitude of the vector sum [tex]\(\vec{A} + \vec{B}\)[/tex] is approximately [tex]\(240.86\)[/tex] meters.