Find the maximum distance between the points `(3sintheta, 0, 0) and (4costheta, 0, 0).`

Answer

To find the maximum distance between the points \( (3\sin\theta, 0, 0) \) and \( (4\cos\theta, 0, 0) \), we can follow these steps:

### Step 1: Identify the points
Let:
- Point A: \( A(3\sin\theta, 0, 0) \)
- Point B: \( B(4\cos\theta, 0, 0) \)

### Step 2: Use the distance formula
The distance \( d \) between two points \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \) in 3D space is given by:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]
For our points:
\[ d = \sqrt{(4\cos\theta - 3\sin\theta)^2 + (0 - 0)^2 + (0 - 0)^2} \]
This simplifies to:
\[ d = \sqrt{(4\cos\theta - 3\sin\theta)^2} \]
Thus, we have:
\[ d = |4\cos\theta - 3\sin\theta| \]

### Step 3: Find the maximum value of \( |4\cos\theta - 3\sin\theta| \)
To maximize \( |4\cos\theta - 3\sin\theta| \), we can use the identity for the maximum value of the expression \( a\cos\theta + b\sin\theta \):
\[ \text{Maximum value} = \sqrt{a^2 + b^2} \]
Here, \( a = 4 \) and \( b = -3 \).

### Step 4: Calculate \( \sqrt{a^2 + b^2} \)
\[ \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]

### Conclusion
Thus, the maximum distance between the points \( (3\sin\theta, 0, 0) \) and \( (4\cos\theta, 0, 0) \) is:
\[ \boxed{5} \]