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 start with the given equation:

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

Let’s denote the magnitude of \(\vec{A}\) and \(\vec{B}\) as \(A\) and \(B\) respectively. Therefore, we can rewrite the equation as:

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

Since \(A = B\), we can denote both magnitudes as \(k\), where \(k = A = B\). Thus, we have:

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

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 \(A\) and \(B\) with \(k\):

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

This simplifies to:

\[ |\vec{A} + \vec{B}| = \sqrt{2k^2 + 2k^2\cos\theta} = \sqrt{2k^2(1 + \cos\theta)} = k\sqrt{2(1 + \cos\theta)} \]

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

\[ k\sqrt{2(1 + \cos\theta)} = k \]

Assuming \(k \neq 0\), we can divide both sides by \(k\):

\[ \sqrt{2(1 + \cos\theta)} = 1 \]

Now, squaring both sides gives:

\[ 2(1 + \cos\theta) = 1 \]

This simplifies to:

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

Subtracting 1 from both sides:

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

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

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

However, since we are looking for the angle between two vectors, we take the acute angle, which is:

\[ \theta = 120^\circ \]

Thus, the angle between \(\vec{A}\) and \(\vec{B}\) is:

\[ \boxed{120^\circ} \]