Find the angle between the planes `2x+y+z-1=0 and 3x+y+2z-2=0`,

Answer

To find the angle between the two planes given by the equations \(2x + y + z - 1 = 0\) and \(3x + y + 2z - 2 = 0\), 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 \(\vec{n} = \langle a, b, c \rangle\).

For the first plane \(2x + y + z - 1 = 0\):
- The coefficients are \(a_1 = 2\), \(b_1 = 1\), \(c_1 = 1\).
- Thus, the normal vector \(\vec{n_1} = \langle 2, 1, 1 \rangle\).

For the second plane \(3x + y + 2z - 2 = 0\):
- The coefficients are \(a_2 = 3\), \(b_2 = 1\), \(c_2 = 2\).
- Thus, the normal vector \(\vec{n_2} = \langle 3, 1, 2 \rangle\).

### Step 2: Use the formula for the angle between two planes
The angle \(\theta\) between two planes can be found using the formula:
\[ \cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|} \]
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 \(\vec{n_1} \cdot \vec{n_2}\)
\[ \vec{n_1} \cdot \vec{n_2} = (2)(3) + (1)(1) + (1)(2) = 6 + 1 + 2 = 9 \]

### Step 4: Calculate the magnitudes of the normal vectors
\[ |\vec{n_1}| = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \]
\[ |\vec{n_2}| = \sqrt{3^2 + 1^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14} \]

### Step 5: Substitute into the cosine formula
Now substituting the values into the cosine formula:
\[ \cos \theta = \frac{|9|}{\sqrt{6} \cdot \sqrt{14}} = \frac{9}{\sqrt{84}} = \frac{9}{2\sqrt{21}} \]

### Step 6: Find the angle \(\theta\)
To find \(\theta\), we take the inverse cosine:
\[ \theta = \cos^{-1}\left(\frac{9}{2\sqrt{21}}\right) \]

Thus, the angle between the two planes is:
\[ \theta = \cos^{-1}\left(\frac{9}{2\sqrt{21}}\right) \]