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: Write the given equations in a standard form
The given equations are:
\[
6x - 2 = 3y + 1 = 2z - 2
\]
We can set each part equal to a parameter \(t\):
\[
6x - 2 = t, \quad 3y + 1 = t, \quad 2z - 2 = t
\]
### Step 2: Express \(x\), \(y\), and \(z\) in terms of \(t\)
From the equations, we can express \(x\), \(y\), and \(z\) as follows:
1. For \(x\):
\[
6x - 2 = t \implies 6x = t + 2 \implies x = \frac{t + 2}{6}
\]
2. For \(y\):
\[
3y + 1 = t \implies 3y = t - 1 \implies y = \frac{t - 1}{3}
\]
3. For \(z\):
\[
2z - 2 = t \implies 2z = t + 2 \implies z = \frac{t + 2}{2}
\]
### Step 3: Identify the direction ratios
The equations can be rewritten in the form:
\[
x = \frac{1}{6}t + \frac{1}{3}, \quad y = \frac{1}{3}t - \frac{1}{3}, \quad z = \frac{1}{2}t + 1
\]
From these equations, we can identify the coefficients of \(t\) which represent the direction ratios of the line:
- The direction ratio for \(x\) is \(\frac{1}{6}\)
- The direction ratio for \(y\) is \(\frac{1}{3}\)
- The direction ratio for \(z\) is \(\frac{1}{2}\)
To express these as integers, we can multiply by a common factor (in this case, 6):
- Direction ratios: \(1, 2, 3\)
### Step 4: Find 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.
From the direction ratios \(1, 2, 3\), the direction vector \(\mathbf{b}\) can be written as:
\[
\mathbf{b} = i + 2j + 3k
\]
### Step 5: Find a point on the line
To find a point on the line, we can set \(t = 0\):
- For \(x\):
\[
x = \frac{0 + 2}{6} = \frac{1}{3}
\]
- For \(y\):
\[
y = \frac{0 - 1}{3} = -\frac{1}{3}
\]
- For \(z\):
\[
z = \frac{0 + 2}{2} = 1
\]
Thus, a point on the line is \(\left(\frac{1}{3}, -\frac{1}{3}, 1\right)\).
### Step 6: Write the vector equation
Now we can write the vector equation:
\[
\mathbf{r} = \left(\frac{1}{3}i - \frac{1}{3}j + 1k\right) + \lambda (i + 2j + 3k)
\]
### Final Answer
1. Direction Ratios: \(1, 2, 3\)
2. Vector Equation:
\[
\mathbf{r} = \left(\frac{1}{3}i - \frac{1}{3}j + 1k\right) + \lambda (i + 2j + 3k)
\]
