The direction cosines of the line which is perpendicular to the lines with direction ratios -1, 2, 2 and 0, 2, 1 are

Options

A. `(2)/(9), (-1)/(9), (-2)/(9)`
B. `(2)/(3), (-1)/(3), (-2)/(3)`
C. `(2)/(9), (-1)/(9), (2)/(9)`
D. `(2)/(3), (-1)/(3), (2)/(3)`

Answer

To find the direction cosines of the line that is perpendicular to the lines with direction ratios (-1, 2, 2) and (0, 2, 1), we can follow these steps:

### Step 1: Define the Direction Ratios
Let the direction ratios of the first line be \( L_1 = -1, M_1 = 2, N_1 = 2 \) and for the second line \( L_2 = 0, M_2 = 2, N_2 = 1 \).

### Step 2: Set Up the Perpendicularity Conditions
Let the direction ratios of the line we are looking for be \( L_3, M_3, N_3 \). For this line to be perpendicular to both given lines, the following conditions must hold:

1. \( L_1 L_3 + M_1 M_3 + N_1 N_3 = 0 \) (for the first line)
2. \( L_2 L_3 + M_2 M_3 + N_2 N_3 = 0 \) (for the second line)

### Step 3: Substitute the Values
Substituting the values into the equations:

1. From the first line:
\[ -1 \cdot L_3 + 2 \cdot M_3 + 2 \cdot N_3 = 0 \]
This simplifies to:
\[ -L_3 + 2M_3 + 2N_3 = 0 \quad \text{(Equation 1)} \]

2. From the second line:
\[ 0 \cdot L_3 + 2 \cdot M_3 + 1 \cdot N_3 = 0 \]
This simplifies to:
\[ 2M_3 + N_3 = 0 \quad \text{(Equation 2)} \]

### Step 4: Solve the Equations
From Equation 2, we can express \( N_3 \) in terms of \( M_3 \):
\[ N_3 = -2M_3 \]

Now, substitute \( N_3 \) into Equation 1:
\[ -L_3 + 2M_3 + 2(-2M_3) = 0 \]
This simplifies to:
\[ -L_3 + 2M_3 - 4M_3 = 0 \]
\[ -L_3 - 2M_3 = 0 \implies L_3 = -2M_3 \]

### Step 5: Express in Terms of a Parameter
Let \( M_3 = k \) (a parameter). Then we have:
\[ L_3 = -2k, \quad M_3 = k, \quad N_3 = -2k \]

### Step 6: Find Direction Cosines
The direction ratios are \( (-2k, k, -2k) \). To find the direction cosines, we need to normalize these ratios:
\[ \text{Magnitude} = \sqrt{(-2k)^2 + k^2 + (-2k)^2} = \sqrt{4k^2 + k^2 + 4k^2} = \sqrt{9k^2} = 3|k| \]

Thus, the direction cosines are:
\[ \frac{-2k}{3|k|}, \quad \frac{k}{3|k|}, \quad \frac{-2k}{3|k|} \]
This simplifies to:
\[ \left(-\frac{2}{3}, \frac{1}{3}, -\frac{2}{3}\right) \]

### Final Answer
The direction cosines of the line that is perpendicular to the given lines are:
\[ \left(-\frac{2}{3}, \frac{1}{3}, -\frac{2}{3}\right) \]