Consider the planes `3x-6y-2z=15a n d2x+y-2z=5.` find the angle between these planes

Answer

To find the angle between the two given planes, we can follow these steps:

### Step 1: Identify the equations of the planes
The equations of the planes are given as:
1. \( 3x - 6y - 2z = 15 \)
2. \( 2x + y - 2z = 5 \)

### Step 2: Write the equations in the standard form
The standard form of a plane is given by \( Ax + By + Cz = D \). Here, we can identify the coefficients:
- For the first plane, \( A_1 = 3, B_1 = -6, C_1 = -2 \)
- For the second plane, \( A_2 = 2, B_2 = 1, C_2 = -2 \)

### Step 3: Use the formula for the angle between two planes
The angle \( \theta \) between two planes can be calculated using the formula:
\[ \cos \theta = \frac{A_1 A_2 + B_1 B_2 + C_1 C_2}{\sqrt{A_1^2 + B_1^2 + C_1^2} \cdot \sqrt{A_2^2 + B_2^2 + C_2^2}} \]

### Step 4: Substitute the coefficients into the formula
Substituting the values we identified:
\[ \cos \theta = \frac{(3)(2) + (-6)(1) + (-2)(-2)}{\sqrt{3^2 + (-6)^2 + (-2)^2} \cdot \sqrt{2^2 + 1^2 + (-2)^2}} \]

### Step 5: Calculate the numerator
Calculating the numerator:
\[ 3 \cdot 2 = 6, \quad -6 \cdot 1 = -6, \quad -2 \cdot -2 = 4 \]
Thus, the numerator becomes:
\[ 6 - 6 + 4 = 4 \]

### Step 6: Calculate the denominator
Calculating the first part of the denominator:
\[ \sqrt{3^2 + (-6)^2 + (-2)^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7 \]
Calculating the second part of the denominator:
\[ \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \]
Thus, the denominator becomes:
\[ 7 \cdot 3 = 21 \]

### Step 7: Combine the results
Now substituting back into the cosine formula:
\[ \cos \theta = \frac{4}{21} \]

### Step 8: Find the angle \( \theta \)
To find \( \theta \), we take the inverse cosine:
\[ \theta = \cos^{-1}\left(\frac{4}{21}\right) \]

### Final Answer
The angle between the two planes is:
\[ \theta = \cos^{-1}\left(\frac{4}{21}\right) \]

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