If three unit vectors are inclined at an angle of `60^@` with each other, then the magnitude of their resultant vector will be

Options

A. (a) zero
B. (b) `sqrt(6)`
C. (c) `sqrt(3)`
D. (d) `sqrt(9)`

Answer

To find the magnitude of the resultant vector when three unit vectors are inclined at an angle of \(60^\circ\) with each other, we can follow these steps:

### Step 1: Understand the Unit Vectors
Let the three unit vectors be \(\vec{A}\), \(\vec{B}\), and \(\vec{C}\). Since they are unit vectors, we have:
\[ |\vec{A}| = |\vec{B}| = |\vec{C}| = 1 \]

### Step 2: Calculate the Dot Products
Since the vectors are inclined at an angle of \(60^\circ\), we can calculate the dot products:
\[ \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos(60^\circ) = 1 \cdot 1 \cdot \frac{1}{2} = \frac{1}{2} \]
Similarly, we have:
\[ \vec{B} \cdot \vec{C} = \frac{1}{2} \quad \text{and} \quad \vec{C} \cdot \vec{A} = \frac{1}{2} \]

### Step 3: Use the Formula for the Magnitude of the Resultant
The magnitude of the resultant vector \(\vec{R} = \vec{A} + \vec{B} + \vec{C}\) can be calculated using the formula:
\[ |\vec{R}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + |\vec{C}|^2 + 2(\vec{A} \cdot \vec{B} + \vec{B} \cdot \vec{C} + \vec{C} \cdot \vec{A}) \]

### Step 4: Substitute the Values
Substituting the known values:
\[ |\vec{R}|^2 = 1^2 + 1^2 + 1^2 + 2\left(\frac{1}{2} + \frac{1}{2} + \frac{1}{2}\right) \]
This simplifies to:
\[ |\vec{R}|^2 = 3 + 2 \cdot \frac{3}{2} = 3 + 3 = 6 \]

### Step 5: Calculate the Magnitude
Taking the square root gives us:
\[ |\vec{R}| = \sqrt{6} \]

### Final Answer
Thus, the magnitude of the resultant vector is:
\[ \sqrt{6} \]

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