The angle between the two vectors `vecA=2hati+3hatj+4hatk` and `vecB=1hati+2hatj-3hatk` will be :

Explanation

To find the angle between the two vectors \(\vec{A} = 2\hat{i} + 3\hat{j} + 4\hat{k}\) and \(\vec{B} = 1\hat{i} + 2\hat{j} - 3\hat{k}\), we will use the formula:

\[ \cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|} \]

### Step 1: Calculate the dot product \(\vec{A} \cdot \vec{B}\)

The dot product of two vectors \(\vec{A} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}\) and \(\vec{B} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}\) is given by:

\[ \vec{A} \cdot \vec{B} = a_1b_1 + a_2b_2 + a_3b_3 \]

For our vectors:

\[ \vec{A} \cdot \vec{B} = (2)(1) + (3)(2) + (4)(-3) \]

Calculating this:

\[ = 2 + 6 - 12 = -4 \]

### Step 2: Calculate the magnitudes of \(\vec{A}\) and \(\vec{B}\)

The magnitude of a vector \(\vec{A} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}\) is given by:

\[ |\vec{A}| = \sqrt{a_1^2 + a_2^2 + a_3^2} \]

For \(\vec{A}\):

\[ |\vec{A}| = \sqrt{2^2 + 3^2 + 4^2} = \sqrt{4 + 9 + 16} = \sqrt{29} \]

For \(\vec{B}\):

\[ |\vec{B}| = \sqrt{1^2 + 2^2 + (-3)^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \]

### Step 3: Substitute into the cosine formula

Now we substitute the dot product and the magnitudes into the cosine formula:

\[ \cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|} = \frac{-4}{\sqrt{29} \cdot \sqrt{14}} \]

### Step 4: Calculate \(\sqrt{29} \cdot \sqrt{14}\)

Calculating the product of the magnitudes:

\[ \sqrt{29} \cdot \sqrt{14} = \sqrt{29 \cdot 14} = \sqrt{406} \]

### Step 5: Final calculation for \(\cos \theta\)

Now we have:

\[ \cos \theta = \frac{-4}{\sqrt{406}} \]

### Step 6: Find \(\theta\)

To find \(\theta\), we take the inverse cosine:

\[ \theta = \cos^{-1}\left(\frac{-4}{\sqrt{406}}\right) \]

### Summary of the solution

The angle \(\theta\) between the vectors \(\vec{A}\) and \(\vec{B}\) is given by:

\[ \theta = \cos^{-1}\left(\frac{-4}{\sqrt{406}}\right) \]