Find equation of angle bisector of plane `x+2y+3z-z=0 and 2x-3y+z+4=0`.

Answer

To find the equation of the angle bisector of the given planes \( P_1: x + 2y + 3z - 1 = 0 \) and \( P_2: 2x - 3y + z + 4 = 0 \), we will follow these steps:

### Step 1: Write the equations of the planes in standard form
The equations of the planes are already given in the standard form:
- Plane 1: \( P_1: x + 2y + 3z - 1 = 0 \)
- Plane 2: \( P_2: 2x - 3y + z + 4 = 0 \)

### Step 2: Identify the normal vectors of the planes
The normal vector of a plane \( Ax + By + Cz + D = 0 \) is given by the coefficients \( (A, B, C) \).
- For plane \( P_1 \): Normal vector \( \vec{n_1} = (1, 2, 3) \)
- For plane \( P_2 \): Normal vector \( \vec{n_2} = (2, -3, 1) \)

### Step 3: Calculate the magnitudes of the normal vectors
The magnitude of a vector \( (A, B, C) \) is given by \( \sqrt{A^2 + B^2 + C^2} \).
- Magnitude of \( \vec{n_1} \):
\[ |\vec{n_1}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \]
- Magnitude of \( \vec{n_2} \):
\[ |\vec{n_2}| = \sqrt{2^2 + (-3)^2 + 1^2} = \sqrt{4 + 9 + 1} = \sqrt{14} \]

### Step 4: Write the equation of the angle bisector
The equation of the angle bisector of two planes can be expressed as:
\[ \frac{P_1}{|\vec{n_1}|} = \pm \frac{P_2}{|\vec{n_2}|} \]
Substituting the values we have:
\[ \frac{x + 2y + 3z - 1}{\sqrt{14}} = \pm \frac{2x - 3y + z + 4}{\sqrt{14}} \]

### Step 5: Simplify the equation
Since the magnitudes are the same, we can simplify the equation by multiplying through by \( \sqrt{14} \):
1. For the positive case:
\[ x + 2y + 3z - 1 = 2x - 3y + z + 4 \]
Rearranging gives:
\[ x - 5y - 2z + 5 = 0 \]

2. For the negative case:
\[ x + 2y + 3z - 1 = - (2x - 3y + z + 4) \]
Rearranging gives:
\[ 3x - y + 4z + 3 = 0 \]

### Final Result
The equations of the angle bisectors of the given planes are:
1. \( x - 5y - 2z + 5 = 0 \)
2. \( 3x - y + 4z + 3 = 0 \)