The cartesian equation of a line are `6x-2=3y+1=2z-2`. Find its direction ratios and also find the vector of the line.

Answer

To solve the problem, we need to find the direction ratios and the vector equation of the line given by the Cartesian equations \(6x - 2 = 3y + 1 = 2z - 2\).

### Step 1: Rewrite the Cartesian equations
We start with the given equation:
\[ 6x - 2 = 3y + 1 = 2z - 2 \]

### Step 2: Set the equations equal to a parameter
Let us set each part of the equation equal to a parameter \(t\):
\[ 6x - 2 = t \]
\[ 3y + 1 = t \]
\[ 2z - 2 = t \]

### Step 3: Solve for \(x\), \(y\), and \(z\)
From the equations, we can express \(x\), \(y\), and \(z\) in terms of \(t\):

1. From \(6x - 2 = t\):
\[ 6x = t + 2 \implies x = \frac{t + 2}{6} \]

2. From \(3y + 1 = t\):
\[ 3y = t - 1 \implies y = \frac{t - 1}{3} \]

3. From \(2z - 2 = t\):
\[ 2z = t + 2 \implies z = \frac{t + 2}{2} \]

### Step 4: Express in symmetric form
We can express the equations in symmetric form:
\[ \frac{x - \frac{1}{3}}{\frac{1}{3}} = \frac{y + \frac{1}{3}}{\frac{2}{3}} = \frac{z - 1}{\frac{3}{2}} \]

### Step 5: Identify direction ratios
From the symmetric form, we can identify the direction ratios \(a\), \(b\), and \(c\):
- The coefficients of \(x\), \(y\), and \(z\) in the symmetric form give us the direction ratios:
- \(a = 1\)
- \(b = 2\)
- \(c = 3\)

Thus, the direction ratios of the line are \(1:2:3\).

### Step 6: Write the vector equation of the line
The vector equation of a line can be expressed as:
\[ \mathbf{r} = \mathbf{a} + \lambda \mathbf{b} \]
where \(\mathbf{a}\) is a position vector of a point on the line and \(\mathbf{b}\) is the direction vector.

1. The point on the line can be taken from the symmetric form, for example, when \(t = 0\):
\[ x = \frac{2}{6} = \frac{1}{3}, \quad y = \frac{-1}{3}, \quad z = 1 \]
Thus, the point is \(\left(\frac{1}{3}, -\frac{1}{3}, 1\right)\).

2. The direction vector \(\mathbf{b}\) corresponding to the direction ratios \(1:2:3\) is:
\[ \mathbf{b} = i + 2j + 3k \]

### Final Vector Equation
Putting it all together, the vector equation of the line is:
\[ \mathbf{r} = \left(\frac{1}{3}i - \frac{1}{3}j + 1k\right) + \lambda (i + 2j + 3k) \]

### Summary
- Direction Ratios: \(1:2:3\)
- Vector Equation:
\[ \mathbf{r} = \left(\frac{1}{3}i - \frac{1}{3}j + 1k\right) + \lambda (i + 2j + 3k) \]