The angle between the planes 2x-y+3z=6 and x+y+2z=7 is

Options

A. `0^@`
B. `30^@`
C. `45^@`
D. `60^@`

Answer

To find the angle between the planes given by the equations \(2x - y + 3z = 6\) 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 represented as \(\vec{n} = ai + bj + ck\).

For the first plane \(2x - y + 3z = 6\):
- The coefficients are \(a = 2\), \(b = -1\), and \(c = 3\).
- Therefore, the normal vector \(\vec{n_1} = 2i - j + 3k\).

For the second plane \(x + y + 2z = 7\):
- The coefficients are \(a = 1\), \(b = 1\), and \(c = 2\).
- Therefore, the normal vector \(\vec{n_2} = i + j + 2k\).

### Step 2: Calculate the dot product of the normal vectors
The dot product of two vectors \(\vec{a} = ai + bj + ck\) and \(\vec{b} = xi + yj + zk\) is given by:
\[ \vec{a} \cdot \vec{b} = ax + by + cz \]

Calculating 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 3: Calculate the magnitudes of the normal vectors
The magnitude of a vector \(\vec{a} = ai + bj + ck\) is given by:
\[ |\vec{a}| = \sqrt{a^2 + b^2 + c^2} \]

Calculating the magnitude of \(\vec{n_1}\):
\[ |\vec{n_1}| = \sqrt{2^2 + (-1)^2 + 3^2} = \sqrt{4 + 1 + 9} = \sqrt{14} \]

Calculating the magnitude of \(\vec{n_2}\):
\[ |\vec{n_2}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6} \]

### Step 4: Use the dot product to find the cosine of the angle
The cosine of the angle \(\theta\) between the two planes (or their normals) can be calculated using the formula:
\[ \cos \theta = \frac{\vec{n_1} \cdot \vec{n_2}}{|\vec{n_1}| |\vec{n_2}|} \]

Substituting the values we calculated:
\[ \cos \theta = \frac{7}{\sqrt{14} \cdot \sqrt{6}} = \frac{7}{\sqrt{84}} = \frac{7}{2\sqrt{21}} \]

### Step 5: Find the angle \(\theta\)
To find the angle \(\theta\), we take the inverse cosine:
\[ \theta = \cos^{-1}\left(\frac{7}{2\sqrt{21}}\right) \]

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