Find the direction cosines of the normal to the plane `2x+3y-z=4`.

Explanation

To find the direction cosines of the normal to the plane given by the equation \(2x + 3y - z = 4\), we can follow these steps:

### Step 1: Identify the normal vector
The equation of the plane can be expressed in the form \(Ax + By + Cz = D\), where \(A\), \(B\), and \(C\) are the coefficients of \(x\), \(y\), and \(z\) respectively. For the given equation \(2x + 3y - z = 4\), we can identify:

- \(A = 2\)
- \(B = 3\)
- \(C = -1\)

Thus, the normal vector \(\mathbf{n}\) to the plane is given by:
\[ \mathbf{n} = \langle 2, 3, -1 \rangle \]

### Step 2: Calculate the magnitude of the normal vector
The magnitude (or length) of the normal vector \(\mathbf{n}\) is calculated using the formula:
\[ |\mathbf{n}| = \sqrt{A^2 + B^2 + C^2} \]
Substituting the values of \(A\), \(B\), and \(C\):
\[ |\mathbf{n}| = \sqrt{2^2 + 3^2 + (-1)^2} = \sqrt{4 + 9 + 1} = \sqrt{14} \]

### Step 3: Find the direction cosines
The direction cosines (\(l\), \(m\), \(n\)) of the normal vector are given by the ratios of the components of the normal vector to its magnitude:
\[ l = \frac{A}{|\mathbf{n}|}, \quad m = \frac{B}{|\mathbf{n}|}, \quad n = \frac{C}{|\mathbf{n}|} \]
Substituting the values:
\[ l = \frac{2}{\sqrt{14}}, \quad m = \frac{3}{\sqrt{14}}, \quad n = \frac{-1}{\sqrt{14}} \]

### Final Answer
Thus, the direction cosines of the normal to the plane \(2x + 3y - z = 4\) are:
\[ \left( \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}, \frac{-1}{\sqrt{14}} \right) \]

---