Find the angle between the pair of lines
`(x-1)/2=(y-3)/4=(z+2)/6 and (x+1)/1=(y-2)/1=(z-3)/4`

Answer

To find the angle between the two given lines, we will follow these steps:

### Step 1: Identify the direction ratios of the lines
The lines are given in symmetric form:
1. For the first line: \((x-1)/2 = (y-3)/4 = (z+2)/6\)
- Direction ratios \( \mathbf{b_1} = (2, 4, 6) \)

2. For the second line: \((x+1)/1 = (y-2)/1 = (z-3)/4\)
- Direction ratios \( \mathbf{b_2} = (1, 1, 4) \)

### Step 2: Use the dot product to find the angle
The formula for the cosine of the angle \( \theta \) between two vectors \( \mathbf{b_1} \) and \( \mathbf{b_2} \) is given by:
\[ \cos \theta = \frac{\mathbf{b_1} \cdot \mathbf{b_2}}{|\mathbf{b_1}| |\mathbf{b_2}|} \]

### Step 3: Calculate the dot product \( \mathbf{b_1} \cdot \mathbf{b_2} \)
\[ \mathbf{b_1} \cdot \mathbf{b_2} = (2)(1) + (4)(1) + (6)(4) = 2 + 4 + 24 = 30 \]

### Step 4: Calculate the magnitudes of \( \mathbf{b_1} \) and \( \mathbf{b_2} \)
- Magnitude of \( \mathbf{b_1} \):
\[ |\mathbf{b_1}| = \sqrt{2^2 + 4^2 + 6^2} = \sqrt{4 + 16 + 36} = \sqrt{56} \]

- Magnitude of \( \mathbf{b_2} \):
\[ |\mathbf{b_2}| = \sqrt{1^2 + 1^2 + 4^2} = \sqrt{1 + 1 + 16} = \sqrt{18} \]

### Step 5: Substitute the values into the cosine formula
\[ \cos \theta = \frac{30}{\sqrt{56} \cdot \sqrt{18}} \]

### Step 6: Simplify the expression
Calculating \( \sqrt{56} \cdot \sqrt{18} = \sqrt{1008} \):
\[ \sqrt{1008} = \sqrt{16 \cdot 63} = 4\sqrt{63} \]
Thus,
\[ \cos \theta = \frac{30}{4\sqrt{63}} = \frac{15}{2\sqrt{63}} \]

### Step 7: Find the angle \( \theta \)
To find \( \theta \), we take the inverse cosine:
\[ \theta = \cos^{-1}\left(\frac{15}{2\sqrt{63}}\right) \]

### Final Answer
The angle between the two given lines is:
\[ \theta = \cos^{-1}\left(\frac{15}{2\sqrt{63}}\right) \]

---