What is the acute angle between the planes `x+y+2z=3 and -2x+y-z=11`?

Options

A. A) `pi/5`
B. B) `pi/4`
C. C) `pi/6`
D. D) `pi/3`

Answer

To find the acute angle between the planes given by the equations \(x + y + 2z = 3\) and \(-2x + y - z = 11\), we can follow these steps:

### Step 1: Write the equations in standard form
The equations of the planes can be rewritten in the standard form \(Ax + By + Cz + D = 0\).

1. For the first plane:
\[ x + y + 2z - 3 = 0 \quad \text{(Equation 1)} \]
Here, \(A_1 = 1\), \(B_1 = 1\), \(C_1 = 2\), and \(D_1 = -3\).

2. For the second plane:
\[ -2x + y - z - 11 = 0 \quad \text{(Equation 2)} \]
Here, \(A_2 = -2\), \(B_2 = 1\), \(C_2 = -1\), and \(D_2 = -11\).

### Step 2: Use the formula for the angle between two planes
The acute angle \(\theta\) between two planes can be found 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 3: Calculate the numerator
Substituting the values into the numerator:
\[ |A_1 A_2 + B_1 B_2 + C_1 C_2| = |(1)(-2) + (1)(1) + (2)(-1)| \]
Calculating this gives:
\[ = |-2 + 1 - 2| = |-3| = 3 \]

### Step 4: Calculate the denominator
Now, we calculate the denominator:
1. For the first plane:
\[ \sqrt{A_1^2 + B_1^2 + C_1^2} = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6} \]

2. For the second plane:
\[ \sqrt{A_2^2 + B_2^2 + C_2^2} = \sqrt{(-2)^2 + 1^2 + (-1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \]

Thus, the denominator becomes:
\[ \sqrt{6} \cdot \sqrt{6} = 6 \]

### Step 5: Calculate \(\cos \theta\)
Now substituting the values into the formula for \(\cos \theta\):
\[ \cos \theta = \frac{3}{6} = \frac{1}{2} \]

### Step 6: Find \(\theta\)
To find the angle \(\theta\):
\[ \theta = \cos^{-1}\left(\frac{1}{2}\right) = 60^\circ \]

### Conclusion
The acute angle between the planes \(x + y + 2z = 3\) and \(-2x + y - z = 11\) is \(60^\circ\).

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