The direction cosines of the line `6x-2=3y+1=2z-2` are

Options

A. `1/(sqrt(3)),1/(sqrt(3)),1/(sqrt(3))`
B. `1/(sqrt(14)),2/(sqrt(14)),3/(sqrt(14))`
C. `1,2,3`
D. none of these

Explanation

To find the direction cosines of the line given by the equation \(6x - 2 = 3y + 1 = 2z - 2\), we will follow these steps:

### Step 1: Rewrite the equation in standard form
The given equation is \(6x - 2 = 3y + 1 = 2z - 2\). We can express this in the form of a parametric equation by introducing a parameter \(t\):
\[ \frac{6x - 2}{6} = \frac{3y + 1}{3} = \frac{2z - 2}{2} = t \]

### Step 2: Express \(x\), \(y\), and \(z\) in terms of \(t\)
From the equations, we can express \(x\), \(y\), and \(z\) as follows:
- From \(6x - 2 = 6t\), we get:
\[ x = t + \frac{1}{3} \]
- From \(3y + 1 = 3t\), we get:
\[ y = t - \frac{1}{3} \]
- From \(2z - 2 = 2t\), we get:
\[ z = t + 1 \]

### Step 3: Identify the direction ratios
The coefficients of \(t\) in the parametric equations give the direction ratios of the line. Thus, the direction ratios are:
\[ (1, 1, 1) \]

### Step 4: Normalize the direction ratios to get direction cosines
To find the direction cosines, we need to convert the direction ratios into a unit vector. The magnitude of the direction ratios is calculated as follows:
\[ \text{Magnitude} = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \]

### Step 5: Calculate the direction cosines
Now we can find the direction cosines by dividing each direction ratio by the magnitude:
\[ \text{Direction cosines} = \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) \]

### Conclusion
The direction cosines of the line are:
\[ \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) \]