Find the angle between the lines given by
`vecr=(2hati+3hatj+4hatk)-lambda (hati-4hatj+5hatk) and vecr = (hati-hatj+hatk)-s(2hati-3hatj+4hatk)`

Answer

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

### Step 1: Identify the direction vectors of the lines
The first line is given by:
\[ \vec{r} = (2\hat{i} + 3\hat{j} + 4\hat{k}) - \lambda(\hat{i} - 4\hat{j} + 5\hat{k}) \]
From this, we can extract the direction vector \(\vec{b_1}\):
\[ \vec{b_1} = \hat{i} - 4\hat{j} + 5\hat{k} \]

The second line is given by:
\[ \vec{r} = (\hat{i} - \hat{j} + \hat{k}) - s(2\hat{i} - 3\hat{j} + 4\hat{k}) \]
From this, we can extract the direction vector \(\vec{b_2}\):
\[ \vec{b_2} = 2\hat{i} - 3\hat{j} + 4\hat{k} \]

### 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 \(\vec{b_1}\) and \(\vec{b_2}\) is given by:
\[ \cos \theta = \frac{\vec{b_1} \cdot \vec{b_2}}{|\vec{b_1}| |\vec{b_2}|} \]

### Step 3: Calculate the dot product \(\vec{b_1} \cdot \vec{b_2}\)
Calculating the dot product:
\[ \vec{b_1} \cdot \vec{b_2} = (1)(2) + (-4)(-3) + (5)(4) = 2 + 12 + 20 = 34 \]

### Step 4: Calculate the magnitudes of \(\vec{b_1}\) and \(\vec{b_2}\)
Calculating the magnitude of \(\vec{b_1}\):
\[ |\vec{b_1}| = \sqrt{(1)^2 + (-4)^2 + (5)^2} = \sqrt{1 + 16 + 25} = \sqrt{42} \]

Calculating the magnitude of \(\vec{b_2}\):
\[ |\vec{b_2}| = \sqrt{(2)^2 + (-3)^2 + (4)^2} = \sqrt{4 + 9 + 16} = \sqrt{29} \]

### Step 5: Substitute values into the cosine formula
Substituting the values into the cosine formula:
\[ \cos \theta = \frac{34}{\sqrt{42} \cdot \sqrt{29}} \]

### Step 6: Find the angle \(\theta\)
To find \(\theta\), take the inverse cosine:
\[ \theta = \cos^{-1} \left( \frac{34}{\sqrt{42} \cdot \sqrt{29}} \right) \]

### Final Result
Thus, the angle between the two lines is:
\[ \theta = \cos^{-1} \left( \frac{34}{\sqrt{42} \cdot \sqrt{29}} \right) \]