Find the angle between the pair of lines
`r=3hat(i)+2hat(j)-4hat(k)+lambda(hat(i)+2hat(j)+2hat(k))`
`r=5hat(i)-4hat(k)+mu(3hat(i)+2hat(j)+6hat(k))`

Answer

To find the angle between the given pair of lines, we can use the formula for the cosine of the angle between two vectors. The lines are given in the vector form, and we can extract the direction vectors from them.

### Step 1: Identify the direction vectors of the lines

The equations of the lines are given as:
1. \( r = 3\hat{i} + 2\hat{j} - 4\hat{k} + \lambda(\hat{i} + 2\hat{j} + 2\hat{k}) \)
2. \( r = 5\hat{i} - 4\hat{k} + \mu(3\hat{i} + 2\hat{j} + 6\hat{k}) \)

From these equations, we can identify the direction vectors:
- For the first line, the direction vector \( \mathbf{b_1} = \hat{i} + 2\hat{j} + 2\hat{k} \)
- For the second line, the direction vector \( \mathbf{b_2} = 3\hat{i} + 2\hat{j} + 6\hat{k} \)

### Step 2: Calculate the dot product of the direction vectors

The dot product \( \mathbf{b_1} \cdot \mathbf{b_2} \) is calculated as follows:
\[ \mathbf{b_1} \cdot \mathbf{b_2} = (1)(3) + (2)(2) + (2)(6) \]
Calculating this gives:
\[ \mathbf{b_1} \cdot \mathbf{b_2} = 3 + 4 + 12 = 19 \]

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

Next, we need to calculate the magnitudes of \( \mathbf{b_1} \) and \( \mathbf{b_2} \):
\[ |\mathbf{b_1}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]
\[ |\mathbf{b_2}| = \sqrt{3^2 + 2^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7 \]

### Step 4: Use the cosine formula to find the angle

Now we can use the formula for the cosine of the angle \( \theta \) between the two vectors:
\[ \cos \theta = \frac{\mathbf{b_1} \cdot \mathbf{b_2}}{|\mathbf{b_1}| |\mathbf{b_2}|} \]
Substituting the values we calculated:
\[ \cos \theta = \frac{19}{3 \times 7} = \frac{19}{21} \]

### Step 5: Calculate the angle \( \theta \)

Finally, to find the angle \( \theta \), we take the inverse cosine:
\[ \theta = \cos^{-1}\left(\frac{19}{21}\right) \]

### Conclusion

The angle between the given pair of lines is \( \theta = \cos^{-1}\left(\frac{19}{21}\right) \).