If `|vec(A)+vec(B)|=|vec(A)|=|vec(B)|` then the angle between `vec(A)` and `vec(B)`is

Options

A. `120^(@)`
B. `60^(@)`
C. `90^(@)`
D. `0^(@)`

Explanation

To solve the problem, we need to analyze the given condition:

\[ |\vec{A} + \vec{B}| = |\vec{A}| = |\vec{B}| \]

Let’s denote the magnitudes of vectors \(\vec{A}\) and \(\vec{B}\) as \(A\) and \(B\) respectively. According to the problem, we have:

\[ |\vec{A}| = |\vec{B}| = A \]

This means that both vectors have the same magnitude. We can rewrite the equation as:

\[ |\vec{A} + \vec{B}| = A \]

Now, we can use the formula for the magnitude of the sum of two vectors:

\[ |\vec{A} + \vec{B}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + 2 |\vec{A}| |\vec{B}| \cos \theta} \]

Substituting the magnitudes:

\[ |\vec{A} + \vec{B}| = \sqrt{A^2 + A^2 + 2 A A \cos \theta} \]

This simplifies to:

\[ |\vec{A} + \vec{B}| = \sqrt{2A^2 + 2A^2 \cos \theta} \]

Since we know that \( |\vec{A} + \vec{B}| = A \), we can set the two expressions equal to each other:

\[ A = \sqrt{2A^2 + 2A^2 \cos \theta} \]

Squaring both sides gives:

\[ A^2 = 2A^2 + 2A^2 \cos \theta \]

Rearranging this equation leads us to:

\[ A^2 - 2A^2 = 2A^2 \cos \theta \]

\[ -A^2 = 2A^2 \cos \theta \]

Dividing both sides by \(A^2\) (assuming \(A \neq 0\)):

\[ -1 = 2 \cos \theta \]

Thus, we find:

\[ \cos \theta = -\frac{1}{2} \]

The angle \(\theta\) that corresponds to \(\cos \theta = -\frac{1}{2}\) is:

\[ \theta = 120^\circ \quad \text{or} \quad \theta = 240^\circ \]

However, in the context of vectors, we typically consider the angle between two vectors to be between \(0^\circ\) and \(180^\circ\). Therefore, the angle between \(\vec{A}\) and \(\vec{B}\) is:

\[ \theta = 120^\circ \]

### Final Answer:
The angle between \(\vec{A}\) and \(\vec{B}\) is \(120^\circ\).