A ray of light is incident on a surface in a direction given by vector `A=2i-2j+k`. The normal to that surface passing through the point of incidence is along the vector `N=j-2k`. The unit vector in the direction of reflected ray is given by `R=aj+bj+ck`. Find three equations in terms of `a,b,c` using which we can find the values of `a,b,& c`

Explanation

To solve the problem, we will derive three equations in terms of \(a\), \(b\), and \(c\) that will help us find the values of these variables.

### Step 1: Normalize the Incident and Normal Vectors
Given the incident vector \(A = 2i - 2j + k\) and the normal vector \(N = j - 2k\), we first need to find the unit vectors for both.

1. **Magnitude of the incident vector \(A\)**:
\[ |A| = \sqrt{(2)^2 + (-2)^2 + (1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \]
The unit vector in the direction of the incident ray \(I\) is:
\[ I = \frac{A}{|A|} = \frac{1}{3}(2i - 2j + k) = \frac{2}{3}i - \frac{2}{3}j + \frac{1}{3}k \]

2. **Magnitude of the normal vector \(N\)**:
\[ |N| = \sqrt{(0)^2 + (1)^2 + (-2)^2} = \sqrt{0 + 1 + 4} = \sqrt{5} \]
The unit vector in the direction of the normal \(N\) is:
\[ N_{unit} = \frac{N}{|N|} = \frac{1}{\sqrt{5}}(0i + 1j - 2k) = \frac{1}{\sqrt{5}}j - \frac{2}{\sqrt{5}}k \]

### Step 2: Use the Law of Reflection
According to the law of reflection, the angle of incidence is equal to the angle of reflection. The relationship between the incident vector \(I\), normal vector \(N\), and reflected vector \(R\) can be expressed as:
\[ R = I - 2(I \cdot N)N \]
where \(R\) is the reflected vector.

### Step 3: Calculate the Dot Product \(I \cdot N\)
We need to compute the dot product \(I \cdot N\):
\[ I \cdot N = \left(\frac{2}{3}i - \frac{2}{3}j + \frac{1}{3}k\right) \cdot \left(0i + 1j - 2k\right) = 0 \cdot \frac{2}{3} + 1 \cdot \left(-\frac{2}{3}\right) + (-2) \cdot \frac{1}{3} = -\frac{2}{3} - \frac{2}{3} = -\frac{4}{3} \]

### Step 4: Find the Reflected Vector \(R\)
Substituting \(I \cdot N\) into the reflection formula:
\[ R = I - 2(-\frac{4}{3})N = I + \frac{8}{3}N \]
Calculating the reflected vector:
\[ R = \left(\frac{2}{3}i - \frac{2}{3}j + \frac{1}{3}k\right) + \frac{8}{3}\left(0i + 1j - 2k\right) = \frac{2}{3}i + \left(-\frac{2}{3} + \frac{8}{3}\right)j + \left(\frac{1}{3} - \frac{16}{3}\right)k \]
This simplifies to:
\[ R = \frac{2}{3}i + 2j - 5k \]

### Step 5: Write \(R\) in Terms of \(a\), \(b\), and \(c\)
We can express the reflected vector \(R\) in terms of \(a\), \(b\), and \(c\):
\[ R = ai + bj + ck \]
Comparing coefficients, we have:
- \(a = \frac{2}{3}\)
- \(b = 2\)
- \(c = -5\)

### Step 6: Formulate the Equations
1. **Magnitude of \(R\)**:
\[ a^2 + b^2 + c^2 = 1 \]
This gives us our first equation.

2. **Dot Product Condition**:
\[ R \cdot N = 0 \quad \text{(perpendicularity condition)} \]
This gives us our second equation.

3. **Coplanarity Condition**:
\[ A \times N \cdot R = 0 \]
This gives us our third equation.

### Summary of Equations
1. \(a^2 + b^2 + c^2 = 1\)
2. \(b - 2c = \frac{4}{3}\)
3. \(3a + 4b + 2c = 0\)