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

Explanation

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 will 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 \(x + y + z + 1 = 0\):
- The 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 \(2x - 2y + 2z + 1 = 0\):
- The 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 planes
The cosine of the angle \(\theta\) between two planes 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 normal vectors and \(|\vec{n_1}|\) and \(|\vec{n_2}|\) are the magnitudes of the normal 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 into the cosine formula
Now substituting the values into the cosine formula:
\[ \cos \theta = \frac{2}{\sqrt{3} \cdot 2\sqrt{3}} = \frac{2}{6} = \frac{1}{3} \]

### Final Answer
Thus, the cosine of the angle between the two planes is:
\[ \cos \theta = \frac{1}{3} \]

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