The acute angle between the planes `2x-y+z=5 and x+y +2z =7` is

Options

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

Answer

To find the acute angle between the planes given by the equations \(2x - y + z = 5\) 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 \(2x - y + z = 5\):
- The coefficients are \(A_1 = 2\), \(B_1 = -1\), \(C_1 = 1\).
- Thus, the normal vector \(\vec{n_1} = (2, -1, 1)\).

For the second plane \(x + y + 2z = 7\):
- The coefficients are \(A_2 = 1\), \(B_2 = 1\), \(C_2 = 2\).
- Thus, the normal vector \(\vec{n_2} = (1, 1, 2)\).

### Step 2: Use the formula for the cosine of the angle between two planes
The cosine of the angle \(\theta\) between two planes can be calculated 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 \(\vec{n_1} \cdot \vec{n_2}\)
\[ \vec{n_1} \cdot \vec{n_2} = (2)(1) + (-1)(1) + (1)(2) = 2 - 1 + 2 = 3 \]

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

### Step 5: Substitute 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 \(\theta\)
Since \(\cos \theta = \frac{1}{2}\), we can find \(\theta\):
\[ \theta = \cos^{-1}\left(\frac{1}{2}\right) = 60^\circ \]

### Final Answer
The acute angle between the planes \(2x - y + z = 5\) and \(x + y + 2z = 7\) is \(60^\circ\).

---