Let ` veca, vecb , vecc` be three vectors of equal magnitude such that the angle between each pair is ` pi/3` . If ` |veca + vecb + vecc| = sqrt6` , then `|veca|=`

Options

A. 2
B. `-1`
C. 1
D. `sqrt6//3`

Explanation

To solve the problem step-by-step, we will follow the mathematical reasoning provided in the video transcript.

### Step 1: Define the Magnitudes
Let the magnitude of each vector \( \vec{a}, \vec{b}, \vec{c} \) be denoted as \( | \vec{a} | = | \vec{b} | = | \vec{c} | = a \).

### Step 2: Use the Given Information
We know that the angle between each pair of vectors is \( \frac{\pi}{3} \). Therefore, we can express the dot products as follows:
- \( \vec{a} \cdot \vec{b} = | \vec{a} | | \vec{b} | \cos\left(\frac{\pi}{3}\right) = a \cdot a \cdot \frac{1}{2} = \frac{a^2}{2} \)
- Similarly, \( \vec{b} \cdot \vec{c} = \frac{a^2}{2} \) and \( \vec{c} \cdot \vec{a} = \frac{a^2}{2} \).

### Step 3: Calculate the Magnitude of the Sum of Vectors
We need to calculate \( | \vec{a} + \vec{b} + \vec{c} | \). The formula for the magnitude of the sum of vectors is:
\[ | \vec{a} + \vec{b} + \vec{c} | = \sqrt{ | \vec{a} |^2 + | \vec{b} |^2 + | \vec{c} |^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) } \]
Substituting the known values:
\[ | \vec{a} + \vec{b} + \vec{c} | = \sqrt{ a^2 + a^2 + a^2 + 2\left(\frac{a^2}{2} + \frac{a^2}{2} + \frac{a^2}{2}\right) } \]

### Step 4: Simplify the Expression
This simplifies to:
\[ | \vec{a} + \vec{b} + \vec{c} | = \sqrt{ 3a^2 + 2\left(\frac{3a^2}{2}\right) } = \sqrt{ 3a^2 + 3a^2 } = \sqrt{ 6a^2 } = \sqrt{6} \cdot |a| \]

### Step 5: Set the Magnitude Equal to Given Value
According to the problem, we have:
\[ \sqrt{6} \cdot |a| = \sqrt{6} \]
Dividing both sides by \( \sqrt{6} \):
\[ |a| = 1 \]

### Conclusion
Thus, the magnitude of vector \( \vec{a} \) is:
\[ | \vec{a} | = 1 \]