Find the direction cosines of the line `(2x+1)/6=(3-2z)/8, y=-3`.

Answer

To find the direction cosines of the line given by the equations \((2x + 1)/6 = (3 - 2z)/8\) and \(y = -3\), we can follow these steps:

### Step 1: Rewrite the equations in a standard form
We start with the equations:
\[ \frac{2x + 1}{6} = \frac{3 - 2z}{8} \quad \text{and} \quad y = -3 \]

### Step 2: Set the equations equal to a parameter \(t\)
Let:
\[ \frac{2x + 1}{6} = t \quad \Rightarrow \quad 2x + 1 = 6t \quad \Rightarrow \quad 2x = 6t - 1 \quad \Rightarrow \quad x = 3t - \frac{1}{2} \]

\[ \frac{3 - 2z}{8} = t \quad \Rightarrow \quad 3 - 2z = 8t \quad \Rightarrow \quad -2z = 8t - 3 \quad \Rightarrow \quad z = -4t + \frac{3}{2} \]

The value of \(y\) is constant:
\[ y = -3 \]

### Step 3: Express \(x\), \(y\), and \(z\) in terms of \(t\)
Now we have:
\[ x = 3t - \frac{1}{2}, \quad y = -3, \quad z = -4t + \frac{3}{2} \]

### Step 4: Identify the direction ratios
From the equations above, we can identify the direction ratios \(l\), \(m\), and \(n\):
- The coefficient of \(t\) in \(x\) is \(3\) (so \(l = 3\)).
- The coefficient of \(t\) in \(y\) is \(0\) (so \(m = 0\)).
- The coefficient of \(t\) in \(z\) is \(-4\) (so \(n = -4\)).

Thus, the direction ratios are:
\[ l = 3, \quad m = 0, \quad n = -4 \]

### Step 5: Calculate the magnitude of the direction ratios
Now we calculate the magnitude of the direction ratios:
\[ \sqrt{l^2 + m^2 + n^2} = \sqrt{3^2 + 0^2 + (-4)^2} = \sqrt{9 + 0 + 16} = \sqrt{25} = 5 \]

### Step 6: Find the direction cosines
The direction cosines \(l', m', n'\) are given by:
\[ l' = \frac{l}{\sqrt{l^2 + m^2 + n^2}} = \frac{3}{5}, \quad m' = \frac{m}{\sqrt{l^2 + m^2 + n^2}} = \frac{0}{5} = 0, \quad n' = \frac{n}{\sqrt{l^2 + m^2 + n^2}} = \frac{-4}{5} \]

Thus, the direction cosines are:
\[ \left(\frac{3}{5}, 0, -\frac{4}{5}\right) \]

### Final Answer
The direction cosines of the line are:
\[ \left(\frac{3}{5}, 0, -\frac{4}{5}\right) \]