A unit vector along the direction `hat(i) + hat(j) + hat(k)` has a magnitude :

Options

A. `sqrt(3)`
B. `sqrt(2)`
C. 1
D. 0

Answer

To find the magnitude of a unit vector along the direction of the vector \(\hat{i} + \hat{j} + \hat{k}\), we can follow these steps:

### Step 1: Identify the vector
The given vector is:
\[ \vec{v} = \hat{i} + \hat{j} + \hat{k} \]

### Step 2: Calculate the magnitude of the vector
The magnitude of a vector \(\vec{v} = a\hat{i} + b\hat{j} + c\hat{k}\) is given by the formula:
\[ |\vec{v}| = \sqrt{a^2 + b^2 + c^2} \]
For our vector \(\vec{v} = 1\hat{i} + 1\hat{j} + 1\hat{k}\), we have \(a = 1\), \(b = 1\), and \(c = 1\).

Now, substituting these values into the magnitude formula:
\[ |\vec{v}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \]

### Step 3: Determine the unit vector
A unit vector \(\hat{u}\) in the direction of \(\vec{v}\) is given by:
\[ \hat{u} = \frac{\vec{v}}{|\vec{v}|} \]
Substituting \(\vec{v}\) and its magnitude:
\[ \hat{u} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}} \]

### Step 4: Find the magnitude of the unit vector
By definition, the magnitude of a unit vector is always:
\[ |\hat{u}| = 1 \]

### Conclusion
Thus, the magnitude of the unit vector along the direction \(\hat{i} + \hat{j} + \hat{k}\) is:
\[ \text{Magnitude of the unit vector} = 1 \]