Let `veca=2veci+3vecj+4veck and vecb=3veci+4vecj+5veck`. Find the angle between them.

Explanation

To find the angle between the vectors \(\vec{a}\) and \(\vec{b}\), we can use the formula for the dot product of two vectors, which relates the dot product to the magnitudes of the vectors and the cosine of the angle between them. The formula is given by:

\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \]

Where:
- \(\vec{a} \cdot \vec{b}\) is the dot product of vectors \(\vec{a}\) and \(\vec{b}\).
- \(|\vec{a}|\) and \(|\vec{b}|\) are the magnitudes of vectors \(\vec{a}\) and \(\vec{b}\).
- \(\theta\) is the angle between the two vectors.

### Step 1: Calculate the dot product \(\vec{a} \cdot \vec{b}\)

Given:
\[ \vec{a} = 2\hat{i} + 3\hat{j} + 4\hat{k} \]
\[ \vec{b} = 3\hat{i} + 4\hat{j} + 5\hat{k} \]

The dot product is calculated as follows:
\[ \vec{a} \cdot \vec{b} = (2)(3) + (3)(4) + (4)(5) \]
\[ = 6 + 12 + 20 = 38 \]

### Step 2: Calculate the magnitudes \(|\vec{a}|\) and \(|\vec{b}|\)

The magnitude of vector \(\vec{a}\) is given by:
\[ |\vec{a}| = \sqrt{(2^2) + (3^2) + (4^2)} = \sqrt{4 + 9 + 16} = \sqrt{29} \]

The magnitude of vector \(\vec{b}\) is given by:
\[ |\vec{b}| = \sqrt{(3^2) + (4^2) + (5^2)} = \sqrt{9 + 16 + 25} = \sqrt{50} \]

### Step 3: Substitute the values into the dot product formula

Now, we substitute the values we found into the dot product formula:
\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \]
\[ 38 = \sqrt{29} \cdot \sqrt{50} \cdot \cos \theta \]

### Step 4: Solve for \(\cos \theta\)

First, calculate \(\sqrt{29} \cdot \sqrt{50}\):
\[ \sqrt{29} \cdot \sqrt{50} = \sqrt{1450} \]

Now, we can express \(\cos \theta\):
\[ \cos \theta = \frac{38}{\sqrt{1450}} \]

### Step 5: Calculate \(\theta\)

Finally, we can find the angle \(\theta\) by taking the inverse cosine:
\[ \theta = \cos^{-1}\left(\frac{38}{\sqrt{1450}}\right) \]

### Summary of the Solution Steps:
1. Calculate the dot product \(\vec{a} \cdot \vec{b} = 38\).
2. Calculate the magnitudes \(|\vec{a}| = \sqrt{29}\) and \(|\vec{b}| = \sqrt{50}\).
3. Substitute into the dot product formula and solve for \(\cos \theta\).
4. Calculate \(\theta = \cos^{-1}\left(\frac{38}{\sqrt{1450}}\right)\).