The following equations represent transverse waves :
`z_(1) = A cos(kx - omegat)`,
`z_(2) = A cos (kx + omegat)`, `z_(3) = A cos (ky - omegat)`
Identify the combination (s) of the waves which will produce (i) standing wave(s), (ii) a wave travelling in the direction making an angle of `45^(@)` with the positive `x` and positive `y` axes. In each case, find the positions at which the resultant intensity is always zero.

Explanation

To solve the problem, we need to analyze the given wave equations and determine the combinations that produce standing waves and waves traveling at a 45-degree angle. Let's break it down step by step.

### Step 1: Identify the Wave Equations
The given wave equations are:
1. \( z_1 = A \cos(kx - \omega t) \)
2. \( z_2 = A \cos(kx + \omega t) \)
3. \( z_3 = A \cos(ky - \omega t) \)

### Step 2: Determine the Combination for Standing Waves
Standing waves are formed by the superposition of two waves traveling in opposite directions.

- **Combination for Standing Waves**:
The waves \( z_1 \) and \( z_2 \) are traveling in opposite directions along the x-axis. Therefore, they will produce a standing wave.

### Step 3: Superimpose the Waves
Using the principle of superposition, we can add \( z_1 \) and \( z_2 \):

\[ z = z_1 + z_2 = A \cos(kx - \omega t) + A \cos(kx + \omega t) \]

Using the trigonometric identity \( \cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \):

\[ z = 2A \cos(kx) \cos(\omega t) \]

### Step 4: Find the Resultant Intensity
The intensity \( I \) is proportional to the square of the amplitude \( R \):

\[ I \propto R^2 = (2A \cos(kx))^2 = 4A^2 \cos^2(kx) \]

For the intensity to be zero:

\[ \cos^2(kx) = 0 \implies \cos(kx) = 0 \]

This occurs when:

\[ kx = \left( n + \frac{1}{2} \right) \pi \quad (n = 0, 1, 2, \ldots) \]

Thus, the positions where the intensity is always zero are:

\[ x = \frac{(2n + 1) \pi}{2k} \quad (n = 0, 1, 2, \ldots) \]

### Step 5: Determine the Combination for Waves Traveling at 45 Degrees
To find a wave traveling at a 45-degree angle, we need to consider the waves \( z_1 \) and \( z_3 \):

- **Combination for 45-degree Wave**:
\( z_1 \) travels in the x-direction, and \( z_3 \) travels in the y-direction. Their combination will yield a wave traveling at a 45-degree angle.

### Step 6: Superimpose the Waves
Using the principle of superposition again:

\[ z = z_1 + z_3 = A \cos(kx - \omega t) + A \cos(ky - \omega t) \]

Using the trigonometric identity:

\[ z = 2A \cos\left(\frac{kx + ky - 2\omega t}{2}\right) \cos\left(\frac{kx - ky}{2}\right) \]

### Step 7: Find the Resultant Intensity
The intensity \( I \) is given by:

\[ I \propto (2A \cos\left(\frac{kx - ky}{2}\right))^2 = 4A^2 \cos^2\left(\frac{kx - ky}{2}\right) \]

For the intensity to be zero:

\[ \cos^2\left(\frac{kx - ky}{2}\right) = 0 \implies \cos\left(\frac{kx - ky}{2}\right) = 0 \]

This occurs when:

\[ \frac{kx - ky}{2} = \left(n + \frac{1}{2}\right) \pi \quad (n = 0, 1, 2, \ldots) \]

Thus, the positions where the intensity is always zero are:

\[ kx - ky = (2n + 1) \pi \quad (n = 0, 1, 2, \ldots) \]

### Summary of Results
1. **Standing Waves**: Formed by \( z_1 \) and \( z_2 \).
- Intensity is zero at \( x = \frac{(2n + 1) \pi}{2k} \).

2. **Waves at 45 Degrees**: Formed by \( z_1 \) and \( z_3 \).
- Intensity is zero at \( kx - ky = (2n + 1) \pi \).