The acute angle between the planes `2x-y+z=6` and `x+y+2z=3` is

Options

A. `45^(@)`
B. `60^(@)`
C. `30^(@)`
D. `75^(@)`

Explanation

To find the acute angle between the planes given by the equations \(2x - y + z = 6\) and \(x + y + 2z = 3\), 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 \((A, B, C)\).

- For the first plane \(2x - y + z = 6\), the normal vector \( \mathbf{n_1} \) is:
\[ \mathbf{n_1} = (2, -1, 1) \]

- For the second plane \(x + y + 2z = 3\), the normal vector \( \mathbf{n_2} \) is:
\[ \mathbf{n_2} = (1, 1, 2) \]

### Step 2: Use the dot product to find the cosine of the angle
The cosine of the angle \( \theta \) between the two normal vectors can be found using the dot product formula:
\[ \cos \theta = \frac{\mathbf{n_1} \cdot \mathbf{n_2}}{|\mathbf{n_1}| |\mathbf{n_2}|} \]

### Step 3: Calculate the dot product \( \mathbf{n_1} \cdot \mathbf{n_2} \)
The dot product is calculated as follows:
\[ \mathbf{n_1} \cdot \mathbf{n_2} = (2)(1) + (-1)(1) + (1)(2) = 2 - 1 + 2 = 3 \]

### Step 4: Calculate the magnitudes of the normal vectors
Now, we calculate the magnitudes of \( \mathbf{n_1} \) and \( \mathbf{n_2} \):
\[ |\mathbf{n_1}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \]
\[ |\mathbf{n_2}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6} \]

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

### Step 6: Find the angle \( \theta \)
To find \( \theta \), we take the inverse cosine:
\[ \theta = \cos^{-1}\left(\frac{1}{2}\right) = 60^\circ \]

### Conclusion
The acute angle between the two planes is \(60^\circ\).