Consider the planes `2x+y+z+4=0, and y-z+4=0` Find the angle between them

Answer

To find the angle between the two planes given by the equations \(2x + y + z + 4 = 0\) and \(y - z + 4 = 0\), 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 \((A, B, C)\).

For the first plane \(2x + y + z + 4 = 0\):
- The coefficients are \(A = 2\), \(B = 1\), and \(C = 1\).
- Thus, the normal vector \(\mathbf{n_1} = (2, 1, 1)\).

For the second plane \(y - z + 4 = 0\):
- The coefficients are \(A = 0\), \(B = 1\), and \(C = -1\).
- Thus, the normal vector \(\mathbf{n_2} = (0, 1, -1)\).

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

### Step 3: Calculate the dot product \(\mathbf{n_1} \cdot \mathbf{n_2}\)
\[ \mathbf{n_1} \cdot \mathbf{n_2} = (2, 1, 1) \cdot (0, 1, -1) = 2 \cdot 0 + 1 \cdot 1 + 1 \cdot (-1) = 0 + 1 - 1 = 0 \]

### Step 4: Calculate the magnitudes of the normal vectors
\[ |\mathbf{n_1}| = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \]

\[ |\mathbf{n_2}| = \sqrt{0^2 + 1^2 + (-1)^2} = \sqrt{0 + 1 + 1} = \sqrt{2} \]

### Step 5: Substitute into the formula
\[ \cos \theta = \frac{|0|}{|\mathbf{n_1}| |\mathbf{n_2}|} = \frac{0}{\sqrt{6} \cdot \sqrt{2}} = 0 \]

### Step 6: Find the angle \(\theta\)
Since \(\cos \theta = 0\), we have:
\[ \theta = 90^\circ \]

### Final Answer
The angle between the two planes is \(90^\circ\).

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