If `|vecA+vecB|=|vecA|=|vecB|` then angle between A and B will be

Options

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

Answer

To find the angle between vectors \(\vec{A}\) and \(\vec{B}\) given that \(|\vec{A} + \vec{B}| = |\vec{A}| = |\vec{B}|\), we can follow these steps:

### Step 1: Set up the equations
Given that \(|\vec{A}| = |\vec{B}| = a\), we can denote:
\[ |\vec{A}| = a \quad \text{and} \quad |\vec{B}| = a \]
We also know:
\[ |\vec{A} + \vec{B}| = a \]

### Step 2: Use the formula for the magnitude of the sum of two vectors
The magnitude of the sum of two vectors can be expressed as:
\[ |\vec{A} + \vec{B}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + 2 |\vec{A}| |\vec{B}| \cos \theta} \]
Substituting the values of \(|\vec{A}|\) and \(|\vec{B}|\):
\[ |\vec{A} + \vec{B}| = \sqrt{a^2 + a^2 + 2a^2 \cos \theta} \]
This simplifies to:
\[ |\vec{A} + \vec{B}| = \sqrt{2a^2(1 + \cos \theta)} \]

### Step 3: Set the magnitudes equal
Since we know that \(|\vec{A} + \vec{B}| = a\), we can set up the equation:
\[ \sqrt{2a^2(1 + \cos \theta)} = a \]

### Step 4: Square both sides
Squaring both sides gives:
\[ 2a^2(1 + \cos \theta) = a^2 \]

### Step 5: Simplify the equation
Dividing both sides by \(a^2\) (assuming \(a \neq 0\)):
\[ 2(1 + \cos \theta) = 1 \]
This simplifies to:
\[ 2 + 2\cos \theta = 1 \]

### Step 6: Solve for \(\cos \theta\)
Rearranging gives:
\[ 2\cos \theta = 1 - 2 \]
\[ 2\cos \theta = -1 \]
\[ \cos \theta = -\frac{1}{2} \]

### Step 7: Find the angle \(\theta\)
The angle \(\theta\) for which \(\cos \theta = -\frac{1}{2}\) is:
\[ \theta = 120^\circ \quad \text{(in the second quadrant)} \]

### Final Answer
Thus, the angle between vectors \(\vec{A}\) and \(\vec{B}\) is:
\[ \theta = 120^\circ \]

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