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. `2sqrt3 (hati - hatj + hatk)`
B. ` 2 sqrt3 (-hati + hatj +hatk)`
C. ` 2sqrt3 (hati + hatj -hatk)`
D. `2sqrt3 (hati + hatj +hatk)`

Answer

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

### Step-by-Step Solution:

1. **Understanding the Direction Cosines**:
Since the vector \(\vec{r}\) is equally inclined to the coordinate axes, the direction cosines \(l\), \(m\), and \(n\) will be equal. Therefore, we can write:
\[ l = m = n \]

2. **Using the Magnitude of the Vector**:
The magnitude of the vector \(\vec{r}\) is given by:
\[ |\vec{r}| = \sqrt{l^2 + m^2 + n^2} \]
Since \(l = m = n\), we can denote them as \(l\). Thus, we have:
\[ |\vec{r}| = \sqrt{3l^2} = \sqrt{3}l \]
Given that \(|\vec{r}| = 6\), we can set up the equation:
\[ \sqrt{3}l = 6 \]

3. **Solving for \(l\)**:
To find \(l\), we can rearrange the equation:
\[ l = \frac{6}{\sqrt{3}} = 2\sqrt{3} \]

4. **Finding the Components of the Vector**:
Since \(l = m = n\), we have:
\[ l = m = n = 2\sqrt{3} \]
Thus, the vector \(\vec{r}\) can be expressed in terms of its components as:
\[ \vec{r} = l \hat{i} + m \hat{j} + n \hat{k} = 2\sqrt{3} \hat{i} + 2\sqrt{3} \hat{j} + 2\sqrt{3} \hat{k} \]

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

### Conclusion:
The required vector \(\vec{r}\) is:
\[ \vec{r} = 2\sqrt{3} (\hat{i} + \hat{j} + \hat{k}) \]

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