If projection of a line on X,Y and Z-axes are 6,2 and 3 respectively then direction cosines of the line is

Options

A. `(6/7,2/7,3/7)`
B. `(7/6,7/2,7/3)`
C. `(6/(11),2/(11),3/(11)`
D. None of these

Explanation

To find the direction cosines of the line given its projections on the X, Y, and Z axes, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Projections**:
The projections of the line on the X, Y, and Z axes are given as:
- Projection on X-axis = 6
- Projection on Y-axis = 2
- Projection on Z-axis = 3

2. **Form the Vector**:
We can represent the line in vector form using its projections. The vector \( \mathbf{b} \) can be expressed as:
\[ \mathbf{b} = 6 \hat{i} + 2 \hat{j} + 3 \hat{k} \]

3. **Calculate the Magnitude of the Vector**:
The magnitude \( |\mathbf{b}| \) of the vector is calculated using the formula:
\[ |\mathbf{b}| = \sqrt{(6)^2 + (2)^2 + (3)^2} \]
Calculating each term:
- \( (6)^2 = 36 \)
- \( (2)^2 = 4 \)
- \( (3)^2 = 9 \)

Adding these values:
\[ |\mathbf{b}| = \sqrt{36 + 4 + 9} = \sqrt{49} = 7 \]

4. **Find the Unit Vector**:
The unit vector \( \mathbf{b}_{\text{cap}} \) in the direction of \( \mathbf{b} \) is given by:
\[ \mathbf{b}_{\text{cap}} = \frac{\mathbf{b}}{|\mathbf{b}|} = \frac{6 \hat{i} + 2 \hat{j} + 3 \hat{k}}{7} \]
This simplifies to:
\[ \mathbf{b}_{\text{cap}} = \frac{6}{7} \hat{i} + \frac{2}{7} \hat{j} + \frac{3}{7} \hat{k} \]

5. **Identify Direction Cosines**:
The direction cosines \( l, m, n \) are the components of the unit vector:
- \( l = \frac{6}{7} \)
- \( m = \frac{2}{7} \)
- \( n = \frac{3}{7} \)

### Final Answer:
The direction cosines of the line are:
\[ \left( \frac{6}{7}, \frac{2}{7}, \frac{3}{7} \right) \]