Find the angle between the lines
`vecr = (hati+hatj)+lambda (hati+hatj+hatk)and vecr=(2hati-hatj)+t(2hati+3hatj+hatk)`

Answer

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

### Step 1: Identify the direction vectors of the lines
The first line is given by:
\[ \vec{r_1} = \hat{i} + \hat{j} + \lambda (\hat{i} + \hat{j} + \hat{k}) \]
From this, we can extract the direction vector \( \vec{v_1} \):
\[ \vec{v_1} = \hat{i} + \hat{j} + \hat{k} \]

The second line is given by:
\[ \vec{r_2} = (2\hat{i} - \hat{j}) + t(2\hat{i} + 3\hat{j} + \hat{k}) \]
From this, we can extract the direction vector \( \vec{v_2} \):
\[ \vec{v_2} = 2\hat{i} + 3\hat{j} + \hat{k} \]

### Step 2: Use the dot product to find the cosine of the angle
The formula for the cosine of the angle \( \theta \) between two vectors \( \vec{v_1} \) and \( \vec{v_2} \) is:
\[ \cos \theta = \frac{\vec{v_1} \cdot \vec{v_2}}{|\vec{v_1}| |\vec{v_2}|} \]

### Step 3: Calculate the dot product \( \vec{v_1} \cdot \vec{v_2} \)
Calculating the dot product:
\[ \vec{v_1} \cdot \vec{v_2} = (1)(2) + (1)(3) + (1)(1) = 2 + 3 + 1 = 6 \]

### Step 4: Calculate the magnitudes of \( \vec{v_1} \) and \( \vec{v_2} \)
Calculating the magnitude of \( \vec{v_1} \):
\[ |\vec{v_1}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \]

Calculating the magnitude of \( \vec{v_2} \):
\[ |\vec{v_2}| = \sqrt{2^2 + 3^2 + 1^2} = \sqrt{4 + 9 + 1} = \sqrt{14} \]

### Step 5: Substitute into the cosine formula
Now substituting the values into the cosine formula:
\[ \cos \theta = \frac{6}{\sqrt{3} \cdot \sqrt{14}} = \frac{6}{\sqrt{42}} \]

### Step 6: Find the angle \( \theta \)
To find \( \theta \), we take the inverse cosine:
\[ \theta = \cos^{-1} \left( \frac{6}{\sqrt{42}} \right) \]

### Final Answer
Thus, the angle between the two lines is:
\[ \theta = \cos^{-1} \left( \frac{6}{\sqrt{42}} \right) \]