Let ` veca and vecb` are two vectors inclined at an angle of ` 60^(@) , If |veca|=|vecb|=2` ,the the angle between ` veca and veca + vecb` is

Options

A. ` 30^(@)`
B. `60^(@)`
C. `45^(@)`
D. none of these

Explanation

To find the angle between the vector \(\vec{a}\) and the vector \(\vec{a} + \vec{b}\), we can follow these steps:

### Step 1: Understand the given information
We know:
- The magnitudes of both vectors are equal: \(|\vec{a}| = |\vec{b}| = 2\)
- The angle between the vectors \(\vec{a}\) and \(\vec{b}\) is \(60^\circ\).

### Step 2: Calculate the magnitude of \(\vec{a} + \vec{b}\)
To find the magnitude of the resultant vector \(\vec{a} + \vec{b}\), we can use the formula:
\[ |\vec{a} + \vec{b}| = \sqrt{|\vec{a}|^2 + |\vec{b}|^2 + 2 |\vec{a}| |\vec{b}| \cos \theta} \]
Substituting the values:
- \(|\vec{a}| = 2\)
- \(|\vec{b}| = 2\)
- \(\theta = 60^\circ\) (where \(\cos 60^\circ = \frac{1}{2}\))

So,
\[ |\vec{a} + \vec{b}| = \sqrt{2^2 + 2^2 + 2 \cdot 2 \cdot 2 \cdot \frac{1}{2}} \]
\[ = \sqrt{4 + 4 + 4} = \sqrt{12} = 2\sqrt{3} \]

### Step 3: Use the dot product to find the angle between \(\vec{a}\) and \(\vec{a} + \vec{b}\)
The dot product formula states:
\[ \vec{a} \cdot (\vec{a} + \vec{b}) = |\vec{a}| |\vec{a} + \vec{b}| \cos \alpha \]
Where \(\alpha\) is the angle between \(\vec{a}\) and \(\vec{a} + \vec{b}\).

Calculating \(\vec{a} \cdot (\vec{a} + \vec{b})\):
\[ \vec{a} \cdot (\vec{a} + \vec{b}) = \vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} = |\vec{a}|^2 + |\vec{a}| |\vec{b}| \cos 60^\circ \]
\[ = 2^2 + 2 \cdot 2 \cdot \frac{1}{2} = 4 + 2 = 6 \]

### Step 4: Substitute into the dot product equation
Now substituting back into the dot product equation:
\[ 6 = 2 \cdot 2\sqrt{3} \cos \alpha \]
\[ 6 = 4\sqrt{3} \cos \alpha \]
\[ \cos \alpha = \frac{6}{4\sqrt{3}} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2} \]

### Step 5: Find the angle \(\alpha\)
The value \(\cos \alpha = \frac{\sqrt{3}}{2}\) corresponds to:
\[ \alpha = 30^\circ \]

### Conclusion
Thus, the angle between \(\vec{a}\) and \(\vec{a} + \vec{b}\) is \(30^\circ\).