Vector `vec(A)` is 2`cm` long and is `60^(@)` above the x-axis in the first quadrant. Vector `vec(B)` is `2 cm` long and is `60^(@)` below the x-axis in the fourth quadrant. The sum `vec(A)+vec(B)` is a vector of magnitudes

Options

A. 2`cm` along positive y-axis
B. 2`cm` along positive x-axis
C. 2`cm` along negative y-axis
D. 2`cm` along negative x-axis

Explanation

To find the magnitude of the vector sum \( \vec{A} + \vec{B} \), we will break down the vectors into their components and then add them.

### Step 1: Identify the components of vector \( \vec{A} \)
Vector \( \vec{A} \) is 2 cm long and makes an angle of \( 60^\circ \) with the x-axis. We can find its components using trigonometric functions:
- \( A_x = A \cos(60^\circ) \)
- \( A_y = A \sin(60^\circ) \)

Given \( A = 2 \, \text{cm} \):
- \( A_x = 2 \cos(60^\circ) = 2 \times \frac{1}{2} = 1 \, \text{cm} \)
- \( A_y = 2 \sin(60^\circ) = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3} \, \text{cm} \)

### Step 2: Identify the components of vector \( \vec{B} \)
Vector \( \vec{B} \) is also 2 cm long but makes an angle of \( -60^\circ \) with the x-axis (below the x-axis). Its components are:
- \( B_x = B \cos(-60^\circ) \)
- \( B_y = B \sin(-60^\circ) \)

Given \( B = 2 \, \text{cm} \):
- \( B_x = 2 \cos(-60^\circ) = 2 \times \frac{1}{2} = 1 \, \text{cm} \)
- \( B_y = 2 \sin(-60^\circ) = 2 \times -\frac{\sqrt{3}}{2} = -\sqrt{3} \, \text{cm} \)

### Step 3: Add the components of \( \vec{A} \) and \( \vec{B} \)
Now, we can find the resultant vector \( \vec{R} = \vec{A} + \vec{B} \) by adding the respective components:
- \( R_x = A_x + B_x = 1 + 1 = 2 \, \text{cm} \)
- \( R_y = A_y + B_y = \sqrt{3} - \sqrt{3} = 0 \, \text{cm} \)

### Step 4: Calculate the magnitude of the resultant vector \( \vec{R} \)
The magnitude of the resultant vector can be calculated using the Pythagorean theorem:
\[ |\vec{R}| = \sqrt{R_x^2 + R_y^2} = \sqrt{(2)^2 + (0)^2} = \sqrt{4} = 2 \, \text{cm} \]

### Final Answer
The magnitude of the vector sum \( \vec{A} + \vec{B} \) is \( 2 \, \text{cm} \).

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