If `vec(r) = (hat(i) + 2hat(j) + 3hat(k)) + lambda(hat(i) - hat(j) + hat(k))` and `vec(r) = (hat(i) + 2hat(j) + 3hat(k)) + mu (hat(i) + hat(j) - hat(k))` are two lines, then find the equation of acute angle bisector of two lines.

Answer

To find the equation of the acute angle bisector of the two lines given in vector form, we will follow these steps:

### Step 1: Identify the Direction Vectors
The two lines are given as:
1. Line L1:
\[ \vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k}) \]
The direction vector \( \vec{a} \) for L1 is:
\[ \vec{a} = \hat{i} - \hat{j} + \hat{k} \]

2. Line L2:
\[ \vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \mu(\hat{i} + \hat{j} - \hat{k}) \]
The direction vector \( \vec{b} \) for L2 is:
\[ \vec{b} = \hat{i} + \hat{j} - \hat{k} \]

### Step 2: Calculate the Dot Product and Magnitudes
Next, we calculate the dot product \( \vec{a} \cdot \vec{b} \) and the magnitudes of \( \vec{a} \) and \( \vec{b} \).

- The dot product:
\[ \vec{a} \cdot \vec{b} = (1)(1) + (-1)(1) + (1)(-1) = 1 - 1 - 1 = -1 \]

- The magnitudes:
\[ |\vec{a}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3} \]
\[ |\vec{b}| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{3} \]

### Step 3: Calculate Cosine of the Angle
Using the dot product and magnitudes, we find \( \cos \theta \):
\[ \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{-1}{\sqrt{3} \cdot \sqrt{3}} = \frac{-1}{3} \]
Since \( \cos \theta \) is negative, the angle \( \theta \) is obtuse.

### Step 4: Determine the Direction Vectors for the Bisectors
The acute angle bisector direction vector \( \vec{u} \) is given by:
\[ \vec{u} = \vec{a} - \vec{b} \]
Calculating \( \vec{u} \):
\[ \vec{u} = (\hat{i} - \hat{j} + \hat{k}) - (\hat{i} + \hat{j} - \hat{k}) = (1 - 1)\hat{i} + (-1 - 1)\hat{j} + (1 + 1)\hat{k} = 0\hat{i} - 2\hat{j} + 2\hat{k} \]
Thus,
\[ \vec{u} = -2\hat{j} + 2\hat{k} \]

### Step 5: Write the Equation of the Acute Angle Bisector
The equation of the acute angle bisector can be expressed as:
\[ \vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \mu \cdot \vec{u} \]
Substituting \( \vec{u} \):
\[ \vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \mu (-2\hat{j} + 2\hat{k}) \]

### Final Equation
Thus, the final equation of the acute angle bisector is:
\[ \vec{r} = \hat{i} + (2 - 2\mu)\hat{j} + (3 + 2\mu)\hat{k} \]