Find the magnitude of two vectors ` vec a` and `vec b` having the same magnitude and such that the angle between them is `60^0` and their scalar product is .

Options

A. `|vec a| =1, |vec b|=2`
B. `|vec a| = |vec b|=1`
C. `|vec a| =3, |vec b|=2`
D. `|vec a| = |vec b|=2`

Answer

To find the magnitude of two vectors \(\vec{a}\) and \(\vec{b}\) that have the same magnitude and an angle of \(60^\circ\) between them, with a scalar product of \(\frac{1}{2}\), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Information:**
- Let the magnitude of both vectors \(\vec{a}\) and \(\vec{b}\) be \(M\).
- The angle between the vectors is \(60^\circ\).
- The scalar product (dot product) of the vectors is given as \(\vec{a} \cdot \vec{b} = \frac{1}{2}\).

2. **Use the Formula for Scalar Product:**
The scalar product of two vectors can be expressed as:
\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos(\theta) \]
Since both vectors have the same magnitude \(M\), we can rewrite this as:
\[ \vec{a} \cdot \vec{b} = M \cdot M \cdot \cos(60^\circ) \]

3. **Substitute the Known Values:**
We know that \(\cos(60^\circ) = \frac{1}{2}\). Therefore, we can substitute this value into the equation:
\[ \vec{a} \cdot \vec{b} = M^2 \cdot \frac{1}{2} \]

4. **Set the Equation Equal to the Given Scalar Product:**
We know from the problem that \(\vec{a} \cdot \vec{b} = \frac{1}{2}\). Thus, we can set up the equation:
\[ M^2 \cdot \frac{1}{2} = \frac{1}{2} \]

5. **Solve for \(M^2\):**
To eliminate \(\frac{1}{2}\) from both sides, we multiply both sides by \(2\):
\[ M^2 = 1 \]

6. **Find the Magnitude \(M\):**
Taking the square root of both sides gives us:
\[ M = \pm 1 \]
Since magnitudes are always positive, we take:
\[ M = 1 \]

7. **Conclusion:**
Therefore, the magnitude of both vectors \(\vec{a}\) and \(\vec{b}\) is \(1\).

### Final Answer:
The magnitude of both vectors \(\vec{a}\) and \(\vec{b}\) is \(1\).

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