To find the direction cosines of the line that is perpendicular to the lines with direction cosines proportional to \( (1, -2, -2) \) and \( (0, 2, 1) \), we can follow these steps:
### Step 1: Identify the direction cosines
The direction cosines of the two lines are given as:
- Line 1: \( \mathbf{d_1} = (1, -2, -2) \)
- Line 2: \( \mathbf{d_2} = (0, 2, 1) \)
### Step 2: Find the cross product
To find a line that is perpendicular to both given lines, we need to calculate the cross product \( \mathbf{d_1} \times \mathbf{d_2} \).
Using the determinant formula for the cross product:
\[
\mathbf{d_1} \times \mathbf{d_2} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
1 & -2 & -2 \\
0 & 2 & 1
\end{vmatrix}
\]
### Step 3: Calculate the determinant
Calculating the determinant, we expand it as follows:
\[
\mathbf{d_1} \times \mathbf{d_2} = \mathbf{i} \begin{vmatrix}
-2 & -2 \\
2 & 1
\end{vmatrix} - \mathbf{j} \begin{vmatrix}
1 & -2 \\
0 & 1
\end{vmatrix} + \mathbf{k} \begin{vmatrix}
1 & -2 \\
0 & 2
\end{vmatrix}
\]
Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix}
-2 & -2 \\
2 & 1
\end{vmatrix} = (-2)(1) - (-2)(2) = -2 + 4 = 2 \)
2. \( \begin{vmatrix}
1 & -2 \\
0 & 1
\end{vmatrix} = (1)(1) - (-2)(0) = 1 \)
3. \( \begin{vmatrix}
1 & -2 \\
0 & 2
\end{vmatrix} = (1)(2) - (-2)(0) = 2 \)
Putting it all together:
\[
\mathbf{d_1} \times \mathbf{d_2} = 2\mathbf{i} - 1\mathbf{j} + 2\mathbf{k} = (2, -1, 2)
\]
### Step 4: Find the magnitude of the resulting vector
Next, we need to find the magnitude of the vector \( (2, -1, 2) \):
\[
\text{Magnitude} = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3
\]
### Step 5: Calculate the direction cosines
To find the direction cosines, we divide each component of the vector by its magnitude:
\[
\text{Direction cosines} = \left( \frac{2}{3}, \frac{-1}{3}, \frac{2}{3} \right)
\]
### Final Answer
The direction cosines of the line that is perpendicular to the given lines are:
\[
\left( \frac{2}{3}, -\frac{1}{3}, \frac{2}{3} \right)
\]
