Find the angle between the planes :
(i) 3 x - 6y - 2z = 7 ` " " ` 2x + y - 2z = 5
(ii) 4 x + 8y + z = 8 `" "` and y + z = 4
(iii) 2 x - y + z =6 `" "` and x + y + 2z = 7 .

Answer

To find the angle between two planes given by their equations, we can use the following formula:

\[ \cos \theta = \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}} \]

where \(a_1, b_1, c_1\) are the coefficients of the first plane and \(a_2, b_2, c_2\) are the coefficients of the second plane.

### (i) For the planes \(3x - 6y - 2z = 7\) and \(2x + y - 2z = 5\):

1. **Identify coefficients**:
- For the first plane: \(a_1 = 3\), \(b_1 = -6\), \(c_1 = -2\)
- For the second plane: \(a_2 = 2\), \(b_2 = 1\), \(c_2 = -2\)

2. **Calculate the dot product**:
\[ a_1 a_2 + b_1 b_2 + c_1 c_2 = (3)(2) + (-6)(1) + (-2)(-2) = 6 - 6 + 4 = 4 \]

3. **Calculate the magnitudes**:
\[ \sqrt{a_1^2 + b_1^2 + c_1^2} = \sqrt{3^2 + (-6)^2 + (-2)^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7 \]
\[ \sqrt{a_2^2 + b_2^2 + c_2^2} = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \]

4. **Substitute into the cosine formula**:
\[ \cos \theta = \frac{4}{7 \cdot 3} = \frac{4}{21} \]

5. **Find the angle**:
\[ \theta = \cos^{-1}\left(\frac{4}{21}\right) \]

### (ii) For the planes \(4x + 8y + z = 8\) and \(y + z = 4\):

1. **Identify coefficients**:
- For the first plane: \(a_1 = 4\), \(b_1 = 8\), \(c_1 = 1\)
- For the second plane: \(a_2 = 0\), \(b_2 = 1\), \(c_2 = 1\)

2. **Calculate the dot product**:
\[ a_1 a_2 + b_1 b_2 + c_1 c_2 = (4)(0) + (8)(1) + (1)(1) = 0 + 8 + 1 = 9 \]

3. **Calculate the magnitudes**:
\[ \sqrt{a_1^2 + b_1^2 + c_1^2} = \sqrt{4^2 + 8^2 + 1^2} = \sqrt{16 + 64 + 1} = \sqrt{81} = 9 \]
\[ \sqrt{a_2^2 + b_2^2 + c_2^2} = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{0 + 1 + 1} = \sqrt{2} \]

4. **Substitute into the cosine formula**:
\[ \cos \theta = \frac{9}{9 \cdot \sqrt{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \]

5. **Find the angle**:
\[ \theta = \cos^{-1}\left(\frac{\sqrt{2}}{2}\right) = 45^\circ \text{ or } \frac{\pi}{4} \]

### (iii) For the planes \(2x - y + z = 6\) and \(x + y + 2z = 7\):

1. **Identify coefficients**:
- For the first plane: \(a_1 = 2\), \(b_1 = -1\), \(c_1 = 1\)
- For the second plane: \(a_2 = 1\), \(b_2 = 1\), \(c_2 = 2\)

2. **Calculate the dot product**:
\[ a_1 a_2 + b_1 b_2 + c_1 c_2 = (2)(1) + (-1)(1) + (1)(2) = 2 - 1 + 2 = 3 \]

3. **Calculate the magnitudes**:
\[ \sqrt{a_1^2 + b_1^2 + c_1^2} = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \]
\[ \sqrt{a_2^2 + b_2^2 + c_2^2} = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6} \]

4. **Substitute into the cosine formula**:
\[ \cos \theta = \frac{3}{\sqrt{6} \cdot \sqrt{6}} = \frac{3}{6} = \frac{1}{2} \]

5. **Find the angle**:
\[ \theta = \cos^{-1}\left(\frac{1}{2}\right) = 60^\circ \text{ or } \frac{\pi}{3} \]

### Summary of Results:
1. Angle between the first pair of planes: \(\theta = \cos^{-1}\left(\frac{4}{21}\right)\)
2. Angle between the second pair of planes: \(\theta = 45^\circ\) or \(\frac{\pi}{4}\)
3. Angle between the third pair of planes: \(\theta = 60^\circ\) or \(\frac{\pi}{3}\)