The angle between the planes `2x-y+z=6` and `x+y+2z=7`, is

Options

A. `pi/6`
B. `pi/4`
C. `pi/3`
D. `pi/2`

Explanation

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

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

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

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

### Step 2: Use the formula to find the angle between the normal 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 \(\vec{n_1} \cdot \vec{n_2}\)
\[ \vec{n_1} \cdot \vec{n_2} = (2)(1) + (-1)(1) + (1)(2) = 2 - 1 + 2 = 3 \]

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

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

### Step 6: Find the angle \(\theta\)
Since \(\cos \theta = \frac{1}{2}\), we can find \(\theta\):
\[ \theta = \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} \]

### Conclusion
The angle between the planes \(2x - y + z = 6\) and \(x + y + 2z = 7\) is \(\frac{\pi}{3}\).

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