Find angle between two lines, one of which has direction ratios `lt`2,2,1`gt` while the other one is obtained by joining the points (3, 1, 4) and (7,2, 12).

Answer

To find the angle between two lines, we need to first identify the direction ratios of both lines and then use the formula for the cosine of the angle between them.

### Step-by-Step Solution:

1. **Identify the Direction Ratios of the First Line:**
The direction ratios of the first line are given as \( \langle 2, 2, 1 \rangle \).

2. **Determine the Direction Ratios of the Second Line:**
The second line is defined by the points \( (3, 1, 4) \) and \( (7, 2, 12) \).
To find the direction ratios, we calculate the differences in coordinates:
- \( x \) direction: \( 7 - 3 = 4 \)
- \( y \) direction: \( 2 - 1 = 1 \)
- \( z \) direction: \( 12 - 4 = 8 \)

Thus, the direction ratios of the second line are \( \langle 4, 1, 8 \rangle \).

3. **Set Up the Vectors:**
Let \( \mathbf{b_1} = \langle 2, 2, 1 \rangle \) and \( \mathbf{b_2} = \langle 4, 1, 8 \rangle \).

4. **Calculate the Dot Product:**
The dot product \( \mathbf{b_1} \cdot \mathbf{b_2} \) is calculated as follows:
\[ \mathbf{b_1} \cdot \mathbf{b_2} = (2)(4) + (2)(1) + (1)(8) = 8 + 2 + 8 = 18 \]

5. **Calculate the Magnitudes of the Vectors:**
- Magnitude of \( \mathbf{b_1} \):
\[ |\mathbf{b_1}| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \]
- Magnitude of \( \mathbf{b_2} \):
\[ |\mathbf{b_2}| = \sqrt{4^2 + 1^2 + 8^2} = \sqrt{16 + 1 + 64} = \sqrt{81} = 9 \]

6. **Use the Cosine Formula:**
The cosine of the angle \( \theta \) between the two lines is given by:
\[ \cos \theta = \frac{\mathbf{b_1} \cdot \mathbf{b_2}}{|\mathbf{b_1}| |\mathbf{b_2}|} \]
Substituting the values we calculated:
\[ \cos \theta = \frac{18}{3 \times 9} = \frac{18}{27} = \frac{2}{3} \]

7. **Find the Angle \( \theta \):**
To find \( \theta \), we take the inverse cosine:
\[ \theta = \cos^{-1}\left(\frac{2}{3}\right) \]

### Final Answer:
The angle between the two lines is \( \theta = \cos^{-1}\left(\frac{2}{3}\right) \).