A vector `vecr` is equally inclined with the coordinates axes. If the tip of `vecr` is in the positive octant and `|vecr|=6`, then `vecr` is

Options

A. `2sqrt(3)(hati-hatj+hatk)`
B. `2sqrt(3)(-hati+hatj+hatk)`
C. `2sqrt(3)(hati+hatj-hatk)`
D. `2sqrt(3)(hati+hatj+hatk)`

Explanation

To solve the problem, we need to find the vector \( \vec{r} \) that is equally inclined with the coordinate axes and has a magnitude of 6.

### Step-by-step Solution:

1. **Understanding the Inclination**:
Since the vector \( \vec{r} \) is equally inclined to the coordinate axes, the direction cosines of the vector with respect to the x, y, and z axes are equal. Let's denote these direction cosines as \( l \), \( m \), and \( n \). Therefore, we have:
\[ l = m = n \]

2. **Using the Property of Direction Cosines**:
The sum of the squares of the direction cosines is equal to 1:
\[ l^2 + m^2 + n^2 = 1 \]
Substituting \( l = m = n \) into the equation gives:
\[ 3l^2 = 1 \implies l^2 = \frac{1}{3} \implies l = \frac{1}{\sqrt{3}} \]

3. **Constructing the Vector**:
The vector \( \vec{r} \) can be expressed in terms of its components along the x, y, and z axes:
\[ \vec{r} = l \hat{i} + m \hat{j} + n \hat{k} = l (\hat{i} + \hat{j} + \hat{k}) = \frac{1}{\sqrt{3}} (\hat{i} + \hat{j} + \hat{k}) \]

4. **Finding the Magnitude**:
We know that the magnitude of the vector \( \vec{r} \) is given as 6:
\[ |\vec{r}| = 6 \]
Therefore, we can express \( \vec{r} \) as:
\[ \vec{r} = 6 \cdot \frac{1}{\sqrt{3}} (\hat{i} + \hat{j} + \hat{k}) = 2\sqrt{3} (\hat{i} + \hat{j} + \hat{k}) \]

5. **Final Expression of the Vector**:
Thus, the vector \( \vec{r} \) is:
\[ \vec{r} = 2\sqrt{3} \hat{i} + 2\sqrt{3} \hat{j} + 2\sqrt{3} \hat{k} \]

### Conclusion:
The vector \( \vec{r} \) that is equally inclined with the coordinate axes and has a magnitude of 6 is:
\[ \vec{r} = 2\sqrt{3} (\hat{i} + \hat{j} + \hat{k}) \]