What is the cosine of angle between the planes `x+y+z+1=0 and 2x-2y+2z+1=0`?

Options

A. `1//2`
B. `1//3`
C. `2//3`
D. `1//4`

Answer

To find the cosine of the angle between the two planes given by the equations \( x + y + z + 1 = 0 \) and \( 2x - 2y + 2z + 1 = 0 \), we can follow these steps:

### Step 1: Identify the normal vectors of the planes
The normal vector of a plane given by the equation \( ax + by + cz + d = 0 \) is \( \vec{n} = (a, b, c) \).

For the first plane \( x + y + z + 1 = 0 \):
- Coefficients are \( a_1 = 1, b_1 = 1, c_1 = 1 \)
- Thus, the normal vector \( \vec{n_1} = (1, 1, 1) \)

For the second plane \( 2x - 2y + 2z + 1 = 0 \):
- Coefficients are \( a_2 = 2, b_2 = -2, c_2 = 2 \)
- Thus, the normal vector \( \vec{n_2} = (2, -2, 2) \)

### Step 2: Use the formula for the cosine of the angle between two vectors
The cosine of the angle \( \theta \) between two vectors \( \vec{n_1} \) and \( \vec{n_2} \) can be calculated using the formula:
\[ \cos \theta = \frac{\vec{n_1} \cdot \vec{n_2}}{|\vec{n_1}| |\vec{n_2}|} \]
where \( \vec{n_1} \cdot \vec{n_2} \) is the dot product of the vectors and \( |\vec{n_1}| \) and \( |\vec{n_2}| \) are the magnitudes of the vectors.

### Step 3: Calculate the dot product \( \vec{n_1} \cdot \vec{n_2} \)
\[ \vec{n_1} \cdot \vec{n_2} = (1)(2) + (1)(-2) + (1)(2) = 2 - 2 + 2 = 2 \]

### Step 4: Calculate the magnitudes of the normal vectors
\[ |\vec{n_1}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \]
\[ |\vec{n_2}| = \sqrt{2^2 + (-2)^2 + 2^2} = \sqrt{4 + 4 + 4} = \sqrt{12} = 2\sqrt{3} \]

### Step 5: Substitute the values into the cosine formula
\[ \cos \theta = \frac{2}{\sqrt{3} \cdot 2\sqrt{3}} = \frac{2}{6} = \frac{1}{3} \]

### Final Answer
The cosine of the angle between the planes is \( \cos \theta = \frac{1}{3} \).

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