Given that A + B = R and A = B = R. What should be the angle between A and B ?

Options

A. `0`
B. `pi//3`
C. `2pi//3`
D. `pi`

Answer

To solve the problem, we need to find the angle between the vectors A and B given the equations A + B = R and A = B = R.

### Step-by-Step Solution:

1. **Understanding the Vectors**:
We have three vectors: A, B, and R. According to the problem, we know that:
\[ A + B = R \]
and
\[ A = B = R. \]

2. **Substituting the Values**:
Since A = B, we can substitute B with A in the first equation:
\[ A + A = R \implies 2A = R. \]

3. **Magnitude of Vectors**:
The magnitudes of the vectors are:
\[ |A| = |B| = |R| = A. \]
Therefore, we can write:
\[ |R| = 2|A| \implies |R| = 2A. \]

4. **Using the Cosine Rule**:
The relationship between the magnitudes of vectors A and B and the angle θ between them can be expressed using the cosine rule:
\[ |R|^2 = |A|^2 + |B|^2 + 2|A||B|\cos(\theta). \]
Since |A| = |B| = A, we have:
\[ |R|^2 = A^2 + A^2 + 2A \cdot A \cdot \cos(\theta). \]
Simplifying this gives:
\[ |R|^2 = 2A^2 + 2A^2\cos(\theta). \]

5. **Substituting the Magnitude of R**:
We know that |R| = 2A, so we can substitute this into the equation:
\[ (2A)^2 = 2A^2 + 2A^2\cos(\theta). \]
This simplifies to:
\[ 4A^2 = 2A^2 + 2A^2\cos(\theta). \]

6. **Solving for Cosine**:
Rearranging the equation gives:
\[ 4A^2 - 2A^2 = 2A^2\cos(\theta) \implies 2A^2 = 2A^2\cos(\theta). \]
Dividing both sides by 2A^2 (assuming A ≠ 0) gives:
\[ 1 = \cos(\theta). \]

7. **Finding the Angle**:
The equation \( \cos(\theta) = 1 \) implies that:
\[ \theta = 0^\circ. \]

### Final Answer:
The angle between vectors A and B is \( 0^\circ \).