If `|baru|=3` and `baru` is equally inclined to the unit vectors `hati, hatj, hatk`, then `baru=`

Options

A. `pm 3(hati+hatj+hatk)`
B. `pm sqrt(3)(hati+hatj+hatk)`
C. `pm(1)/(3)(hati+hatj+hatk)`
D. `pm(1)/(sqrt(3))(hati+hatj+hatk)`

Answer

To solve the problem, we need to find the vector \( \mathbf{u} \) given that its magnitude is 3 and it is equally inclined to the unit vectors \( \hat{i}, \hat{j}, \hat{k} \).

### Step-by-Step Solution:

1. **Understanding the Inclination**:
Since \( \mathbf{u} \) is equally inclined to the unit vectors \( \hat{i}, \hat{j}, \hat{k} \), it means that the direction cosines of \( \mathbf{u} \) with respect to the coordinate axes are equal. Let these direction cosines be \( l, m, n \). Therefore, we have:
\[ l = m = n \]

2. **Using the Direction Cosine Relation**:
The relation between the direction cosines is given by:
\[ l^2 + m^2 + n^2 = 1 \]
Substituting \( l = m = n \) into this equation gives:
\[ 3l^2 = 1 \]
From this, we can solve for \( l \):
\[ l^2 = \frac{1}{3} \quad \Rightarrow \quad l = \pm \frac{1}{\sqrt{3}} \]

3. **Finding the Components of \( \mathbf{u} \)**:
Since \( l = m = n \), we have:
\[ l = m = n = \pm \frac{1}{\sqrt{3}} \]

4. **Expressing the Vector \( \mathbf{u} \)**:
The vector \( \mathbf{u} \) can be expressed in terms of its magnitude and direction cosines:
\[ \mathbf{u} = |u| (l \hat{i} + m \hat{j} + n \hat{k}) \]
Given that \( |u| = 3 \), we substitute the values of \( l, m, n \):
\[ \mathbf{u} = 3 \left( \pm \frac{1}{\sqrt{3}} \hat{i} + \pm \frac{1}{\sqrt{3}} \hat{j} + \pm \frac{1}{\sqrt{3}} \hat{k} \right) \]

5. **Simplifying the Expression**:
We can factor out \( \pm \frac{1}{\sqrt{3}} \):
\[ \mathbf{u} = \pm \frac{3}{\sqrt{3}} (\hat{i} + \hat{j} + \hat{k}) \]
Simplifying \( \frac{3}{\sqrt{3}} \):
\[ \mathbf{u} = \pm \sqrt{3} (\hat{i} + \hat{j} + \hat{k}) \]

### Final Answer:
Thus, the vector \( \mathbf{u} \) is:
\[ \mathbf{u} = \pm \sqrt{3} (\hat{i} + \hat{j} + \hat{k}) \]