The projection of a vector on the three coordinate axes are `6, -3, 2`, respectively. The direction cosines of the vector are

Options

A. `6, -3, 2`
B. `(6)/(5), -(3)/(5), (2)/(5)`
C. `(6)/(7), -(3)/(7), (2)/(7)`
D. `-(6)/(7), -(3)/(7), (2)/(7)`

Answer

To find the direction cosines of the vector given its projections on the coordinate axes, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Projections**:
The projections of the vector on the x, y, and z axes are given as:
- \( P_x = 6 \)
- \( P_y = -3 \)
- \( P_z = 2 \)

2. **Use the Relationship Between Projections and Direction Cosines**:
The relationship between the projections and the direction cosines is given by:
\[ r \cos \alpha = P_x, \quad r \cos \beta = P_y, \quad r \cos \gamma = P_z \]
where \( r \) is the magnitude of the vector, and \( \alpha, \beta, \gamma \) are the angles the vector makes with the x, y, and z axes respectively.

3. **Square the Projections**:
We can express the equations based on the projections:
\[ r^2 \cos^2 \alpha = 6^2 = 36 \quad (1) \]
\[ r^2 \cos^2 \beta = (-3)^2 = 9 \quad (2) \]
\[ r^2 \cos^2 \gamma = 2^2 = 4 \quad (3) \]

4. **Add the Squared Equations**:
Adding equations (1), (2), and (3):
\[ r^2 \cos^2 \alpha + r^2 \cos^2 \beta + r^2 \cos^2 \gamma = 36 + 9 + 4 = 49 \]
This simplifies to:
\[ r^2 (\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma) = 49 \]

5. **Use the Property of Direction Cosines**:
We know that:
\[ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \]
Therefore, substituting this into the equation gives:
\[ r^2 \cdot 1 = 49 \implies r^2 = 49 \]
Thus,
\[ r = 7 \]

6. **Calculate the Direction Cosines**:
Now, we can find the direction cosines:
\[ \cos \alpha = \frac{P_x}{r} = \frac{6}{7}, \quad \cos \beta = \frac{P_y}{r} = \frac{-3}{7}, \quad \cos \gamma = \frac{P_z}{r} = \frac{2}{7} \]

7. **Final Result**:
The direction cosines of the vector are:
\[ \left( \frac{6}{7}, \frac{-3}{7}, \frac{2}{7} \right) \]

### Conclusion:
The correct option for the direction cosines is:
\[ \boxed{\left( \frac{6}{7}, \frac{-3}{7}, \frac{2}{7} \right)} \]