The angle between the lines
`vecr=(hati+hatj+hatk)+lamda(hati+hatj+2hatk)`
and `vecr=(hati+hatj+hatk)+mu{(-sqrt(3)-1)hati+(sqrt(3)-1)hatj+4hatk}` is

Options

A. `(pi)/6`
B. `(pi)/4`
C. `(pi)/3`
D. `(2pi)/3`

Explanation

To find the angle between the two given lines in vector form, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Direction Vectors**:
The given lines are:
- Line 1: \(\vec{r} = \hat{i} + \hat{j} + \hat{k} + \lambda(\hat{i} + \hat{j} + 2\hat{k})\)
- Line 2: \(\vec{r} = \hat{i} + \hat{j} + \hat{k} + \mu\{(-\sqrt{3}-1)\hat{i} + (\sqrt{3}-1)\hat{j} + 4\hat{k}\}\)

From these equations, we can extract the direction vectors:
- For Line 1, the direction vector \(\vec{b_1} = \hat{i} + \hat{j} + 2\hat{k}\)
- For Line 2, the direction vector \(\vec{b_2} = (-\sqrt{3}-1)\hat{i} + (\sqrt{3}-1)\hat{j} + 4\hat{k}\)

2. **Calculate the Dot Product**:
The dot product \(\vec{b_1} \cdot \vec{b_2}\) is calculated as follows:
\[ \vec{b_1} \cdot \vec{b_2} = (1)(-\sqrt{3}-1) + (1)(\sqrt{3}-1) + (2)(4) \]
Simplifying this:
\[ = -\sqrt{3} - 1 + \sqrt{3} - 1 + 8 = 6 \]

3. **Calculate the Magnitudes of the Direction Vectors**:
- For \(\vec{b_1}\):
\[ |\vec{b_1}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6} \]
- For \(\vec{b_2}\):
\[ |\vec{b_2}| = \sqrt{(-\sqrt{3}-1)^2 + (\sqrt{3}-1)^2 + 4^2} \]
Expanding this:
\[ = \sqrt{(3 + 2\sqrt{3} + 1) + (3 - 2\sqrt{3} + 1) + 16} = \sqrt{(3 + 1 + 3 + 1 + 16)} = \sqrt{24} \]

4. **Use the Dot Product to Find the Cosine of the Angle**:
Using the formula:
\[ \cos \theta = \frac{\vec{b_1} \cdot \vec{b_2}}{|\vec{b_1}| |\vec{b_2}|} \]
Substituting the values:
\[ \cos \theta = \frac{6}{\sqrt{6} \cdot \sqrt{24}} = \frac{6}{\sqrt{144}} = \frac{6}{12} = \frac{1}{2} \]

5. **Calculate the Angle**:
Since \(\cos \theta = \frac{1}{2}\), we find:
\[ \theta = \cos^{-1}\left(\frac{1}{2}\right) = 60^\circ \text{ or } \frac{\pi}{3} \text{ radians} \]

### Final Answer:
The angle between the lines is \(60^\circ\) or \(\frac{\pi}{3}\) radians.

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