What are the direction-cosines of the normal to the plane 3x + 2y - 3z = 8 ?

Answer

To find the direction cosines of the normal to the plane given by the equation \(3x + 2y - 3z = 8\), we can follow these steps:

### Step 1: Identify the coefficients from the plane equation
The general form of the equation of a plane is given by:
\[ ax + by + cz = d \]
where \(a\), \(b\), and \(c\) are the coefficients corresponding to \(x\), \(y\), and \(z\) respectively.

In our case, the equation is:
\[ 3x + 2y - 3z = 8 \]
From this, we can identify:
- \(a = 3\)
- \(b = 2\)
- \(c = -3\)

### Step 2: Determine the direction ratios
The direction ratios of the normal to the plane are simply the coefficients \(a\), \(b\), and \(c\). Therefore, the direction ratios are:
\[ (3, 2, -3) \]

### Step 3: Calculate the magnitude of the direction ratios
The magnitude \(R\) of the direction ratios can be calculated using the formula:
\[ R = \sqrt{a^2 + b^2 + c^2} \]
Substituting the values we have:
\[ R = \sqrt{3^2 + 2^2 + (-3)^2} = \sqrt{9 + 4 + 9} = \sqrt{22} \]

### Step 4: Find the direction cosines
The direction cosines \(l\), \(m\), and \(n\) can be found using the formulas:
\[ l = \frac{a}{R}, \quad m = \frac{b}{R}, \quad n = \frac{c}{R} \]
Substituting the values:
- For \(l\):
\[ l = \frac{3}{\sqrt{22}} \]
- For \(m\):
\[ m = \frac{2}{\sqrt{22}} \]
- For \(n\):
\[ n = \frac{-3}{\sqrt{22}} \]

### Final Answer
Thus, the direction cosines of the normal to the plane \(3x + 2y - 3z = 8\) are:
\[ \left(\frac{3}{\sqrt{22}}, \frac{2}{\sqrt{22}}, \frac{-3}{\sqrt{22}}\right) \]