A plane which bisects the angle between the two given planes `2x - y +2z - 4 = 0` and `x +2y +2z - 2 = 0`, passes through the point: (A) `(1,-4,1)` (B) `(1,4,-1)` (C) `(2,4,1)` (D) `(2,-4,1)`

Options

A. `(1,-4,1)`
B. `(1,4,-1)`
C. `(2,4,1)`
D. `(2,-4,1)`

Answer

To find the plane that bisects the angle between the two given planes \(2x - y + 2z - 4 = 0\) and \(x + 2y + 2z - 2 = 0\), we can use the formula for the angle bisector of two planes.

### Step 1: Identify the coefficients of the planes

For the first plane \(2x - y + 2z - 4 = 0\):
- Coefficients: \(a_1 = 2\), \(b_1 = -1\), \(c_1 = 2\), \(d_1 = -4\)

For the second plane \(x + 2y + 2z - 2 = 0\):
- Coefficients: \(a_2 = 1\), \(b_2 = 2\), \(c_2 = 2\), \(d_2 = -2\)

### Step 2: Write the angle bisector equation

The equation of the plane that bisects the angle between the two planes can be expressed as:

\[ \frac{a_1 x + b_1 y + c_1 z + d_1}{\sqrt{a_1^2 + b_1^2 + c_1^2}} = \pm \frac{a_2 x + b_2 y + c_2 z + d_2}{\sqrt{a_2^2 + b_2^2 + c_2^2}} \]

### Step 3: Calculate the magnitudes of the normal vectors

Calculate the magnitudes of the normal vectors for both planes:

1. For the first plane:
\[ \sqrt{a_1^2 + b_1^2 + c_1^2} = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \]

2. For the second plane:
\[ \sqrt{a_2^2 + b_2^2 + c_2^2} = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]

### Step 4: Set up the angle bisector equation

Substituting the values into the angle bisector equation gives:

\[ \frac{2x - y + 2z - 4}{3} = \pm \frac{x + 2y + 2z - 2}{3} \]

This simplifies to two equations:

1. \(2x - y + 2z - 4 = x + 2y + 2z - 2\)
2. \(2x - y + 2z - 4 = - (x + 2y + 2z - 2)\)

### Step 5: Solve the first equation

From the first equation:

\[ 2x - y + 2z - 4 = x + 2y + 2z - 2 \]

Rearranging gives:

\[ 2x - x - y - 2y - 4 + 2 = 0 \]
\[ x - 3y - 2 = 0 \quad \text{(Equation 1)} \]

### Step 6: Solve the second equation

From the second equation:

\[ 2x - y + 2z - 4 = -x - 2y - 2z + 2 \]

Rearranging gives:

\[ 2x + x - y + 2y + 2z - 2z - 4 - 2 = 0 \]
\[ 3x + y - 6 = 0 \quad \text{(Equation 2)} \]

### Step 7: Find the intersection of the two equations

Now we have two equations:
1. \(x - 3y - 2 = 0\)
2. \(3x + y - 6 = 0\)

We can solve these equations simultaneously.

From Equation 1:
\[ x = 3y + 2 \]

Substituting into Equation 2:
\[ 3(3y + 2) + y - 6 = 0 \]
\[ 9y + 6 + y - 6 = 0 \]
\[ 10y = 0 \implies y = 0 \]

Substituting \(y = 0\) back into Equation 1:
\[ x - 3(0) - 2 = 0 \implies x = 2 \]

### Step 8: Find \(z\)

Substituting \(x = 2\) and \(y = 0\) into either equation to find \(z\):
Using Equation 1:
\[ 2 - 3(0) - 2 = 0 \implies z \text{ can be any value} \]

Thus, the bisector plane is \(z = k\) where \(k\) can take any value.

### Step 9: Check which point lies on the bisector plane

Now we check the provided points to see which one lies on the bisector plane:

1. **Point A**: \((1, -4, 1)\)
- Check: \(2 - 3(-4) - 2 = 0\) (does not satisfy)

2. **Point B**: \((1, 4, -1)\)
- Check: \(1 - 3(4) - 2 = 0\) (does not satisfy)

3. **Point C**: \((2, 4, 1)\)
- Check: \(2 - 3(4) - 2 = 0\) (does not satisfy)

4. **Point D**: \((2, -4, 1)\)
- Check: \(2 - 3(-4) - 2 = 0\) (satisfies)

Thus, the point that lies on the bisector plane is **Point D**: \((2, -4, 1)\).

### Final Answer:
The plane that bisects the angle between the two given planes passes through the point **(2, -4, 1)**.