Find the angle between the planes `2x - y + 3z = 6 and x + y +2z =7`.

Answer

To find the angle between the planes given by the equations \(2x - y + 3z = 6\) and \(x + y + 2z = 7\), we will follow these steps:

### Step 1: Identify the normal vectors of the planes
The normal vector of a plane given by the equation \(Ax + By + Cz = D\) is \(\vec{n} = (A, B, C)\).

For the first plane \(2x - y + 3z = 6\):
- The coefficients are \(A = 2\), \(B = -1\), \(C = 3\).
- Thus, the normal vector \(\vec{n_1} = (2, -1, 3)\).

For the second plane \(x + y + 2z = 7\):
- The coefficients are \(A = 1\), \(B = 1\), \(C = 2\).
- Thus, the normal vector \(\vec{n_2} = (1, 1, 2)\).

### Step 2: Use the formula for the angle between two vectors
The angle \(\theta\) between two vectors \(\vec{n_1}\) and \(\vec{n_2}\) can be found using the dot product formula:
\[ \cos \theta = \frac{\vec{n_1} \cdot \vec{n_2}}{|\vec{n_1}| |\vec{n_2}|} \]

### Step 3: Calculate the dot product \(\vec{n_1} \cdot \vec{n_2}\)
\[ \vec{n_1} \cdot \vec{n_2} = (2)(1) + (-1)(1) + (3)(2) = 2 - 1 + 6 = 7 \]

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

### Step 5: Substitute into the cosine formula
\[ \cos \theta = \frac{7}{|\vec{n_1}| |\vec{n_2}|} = \frac{7}{\sqrt{14} \cdot \sqrt{6}} = \frac{7}{\sqrt{84}} = \frac{7}{2\sqrt{21}} \]

### Step 6: Find the angle \(\theta\)
To find \(\theta\), we take the inverse cosine:
\[ \theta = \cos^{-1}\left(\frac{7}{2\sqrt{21}}\right) \]

Thus, the angle between the planes \(2x - y + 3z = 6\) and \(x + y + 2z = 7\) is:
\[ \theta = \cos^{-1}\left(\frac{7}{2\sqrt{21}}\right) \]