Let's find the magnitude of the resultant vector [tex]\( \mathbf{A} + \mathbf{B} \)[/tex]. Given:
- Length of vector [tex]\( \mathbf{A} \)[/tex], [tex]\( A = 2 \)[/tex] units
- Length of vector [tex]\( \mathbf{B} \)[/tex], [tex]\( B = 2 \)[/tex] units
- The angle between [tex]\( \mathbf{A} \)[/tex] and [tex]\( \mathbf{B} \)[/tex] is [tex]\( 120^\circ \)[/tex]
Here are the steps to solve the problem:
1. Convert the Angle to Radians: The angle given is [tex]\( 120^\circ \)[/tex]. Angles in vectors calculations are generally converted to radians for ease of use.
[tex]\[
\text{Angle in radians} = \frac{120 \times \pi}{180} = \frac{2\pi}{3} \approx 2.0944 \text{ radians}
\][/tex]
2. Use the Law of Cosines to find the magnitude of the resultant vector [tex]\( \mathbf{A} + \mathbf{B} \)[/tex]:
[tex]\[
C^2 = A^2 + B^2 + 2 \cdot A \cdot B \cdot \cos(\theta)
\][/tex]
Where:
- [tex]\( C \)[/tex] is the magnitude of [tex]\( \mathbf{A} + \mathbf{B} \)[/tex]
- [tex]\( A = 2 \)[/tex]
- [tex]\( B = 2 \)[/tex]
- [tex]\( \theta = 120^\circ \)[/tex]
3. Substitute the Values:
[tex]\[
C^2 = 2^2 + 2^2 + 2 \cdot 2 \cdot 2 \cdot \cos(120^\circ)
\][/tex]
Now, using [tex]\( \cos(120^\circ) = -0.5 \)[/tex]:
4. Calculate:
[tex]\[
C^2 = 4 + 4 + 2 \cdot 2 \cdot (-0.5)
\][/tex]
[tex]\[
C^2 = 4 + 4 - 4
\][/tex]
[tex]\[
C^2 = 4
\][/tex]
5. Find the Magnitude:
[tex]\[
C = \sqrt{4} = 2
\][/tex]
Therefore, the magnitude of [tex]\( \mathbf{A} + \mathbf{B} \)[/tex] is [tex]\( 2 \)[/tex] units.
So, the correct answer is:
c. 2
