What is the angle between the plane `2x-y-2z+1=0 and 3x-4y+5z-3=0`?

Options

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

Answer

To find the angle between the two planes given by the equations \(2x - y - 2z + 1 = 0\) and \(3x - 4y + 5z - 3 = 0\), we can follow these steps:

### Step 1: Identify the normal vectors of the planes
The normal vector of a plane given by the equation \(Ax + By + Cz + D = 0\) is \(\vec{n} = (A, B, C)\).

For the first plane \(2x - y - 2z + 1 = 0\):
- The coefficients are \(A_1 = 2\), \(B_1 = -1\), \(C_1 = -2\).
- Thus, the normal vector \(\vec{n_1} = (2, -1, -2)\).

For the second plane \(3x - 4y + 5z - 3 = 0\):
- The coefficients are \(A_2 = 3\), \(B_2 = -4\), \(C_2 = 5\).
- Thus, the normal vector \(\vec{n_2} = (3, -4, 5)\).

### Step 2: Use the formula for the angle between two planes
The angle \(\theta\) between two planes can be found 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 \(\vec{n_1} \cdot \vec{n_2}\)
\[ \vec{n_1} \cdot \vec{n_2} = (2)(3) + (-1)(-4) + (-2)(5) = 6 + 4 - 10 = 0 \]

### Step 4: Calculate the magnitudes of the normal vectors
\[ |\vec{n_1}| = \sqrt{2^2 + (-1)^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \]
\[ |\vec{n_2}| = \sqrt{3^2 + (-4)^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2} \]

### Step 5: Substitute into the cosine formula
\[ \cos \theta = \frac{0}{3 \cdot 5\sqrt{2}} = 0 \]

### Step 6: Find the angle \(\theta\)
Since \(\cos \theta = 0\), this implies:
\[ \theta = \cos^{-1}(0) = 90^\circ \]

### Conclusion
The angle between the two planes is \(90^\circ\) or \(\frac{\pi}{2}\) radians.

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