The angle between two planes `x+2y+2z=3` and `-5x+3y+4z=9` is

Options

A. `cos^(-1).(0sqrt(2))/(20)`
B. `cos^(-1).(3sqrt(2))/(5)`
C. `cos^(-1).(3sqrt(2))/(10)`
D. `cos^(-1).(19sqrt(2))/(30)`

Explanation

To find the angle between the two planes given by the equations \(x + 2y + 2z = 3\) and \(-5x + 3y + 4z = 9\), 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 \(x + 2y + 2z = 3\):
- The coefficients are \(A_1 = 1\), \(B_1 = 2\), \(C_1 = 2\).
- Thus, the normal vector \(\vec{n_1} = (1, 2, 2)\).

For the second plane \(-5x + 3y + 4z = 9\):
- The coefficients are \(A_2 = -5\), \(B_2 = 3\), \(C_2 = 4\).
- Thus, the normal vector \(\vec{n_2} = (-5, 3, 4)\).

### Step 2: Calculate the dot product of the normal vectors
The dot product \(\vec{n_1} \cdot \vec{n_2}\) is calculated as follows:
\[ \vec{n_1} \cdot \vec{n_2} = (1)(-5) + (2)(3) + (2)(4) = -5 + 6 + 8 = 9. \]

### Step 3: Calculate the magnitudes of the normal vectors
The magnitude of a vector \(\vec{n} = (A, B, C)\) is given by \(\|\vec{n}\| = \sqrt{A^2 + B^2 + C^2}\).

For \(\vec{n_1} = (1, 2, 2)\):
\[ \|\vec{n_1}\| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3. \]

For \(\vec{n_2} = (-5, 3, 4)\):
\[ \|\vec{n_2}\| = \sqrt{(-5)^2 + 3^2 + 4^2} = \sqrt{25 + 9 + 16} = \sqrt{50} = 5\sqrt{2}. \]

### Step 4: Use the formula for the cosine of the angle between the planes
The cosine of the angle \(\theta\) between the two planes 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 found:
\[ \cos \theta = \frac{9}{3 \cdot 5\sqrt{2}} = \frac{9}{15\sqrt{2}} = \frac{3}{5\sqrt{2}}. \]

### Step 5: Calculate the angle \(\theta\)
To find the angle \(\theta\), we take the inverse cosine:
\[ \theta = \cos^{-1}\left(\frac{3}{5\sqrt{2}}\right). \]

### Final Answer
Thus, the angle between the two planes is:
\[ \theta = \cos^{-1}\left(\frac{3\sqrt{2}}{10}\right). \]