Find the angle between the following vectors :
(i) `veca = 2hati-6hatj+3hatk` and `vecb = hati+2hatj-2hatk`
(ii) `veca=6hati+3hatj-2hatk` and `vecb=4hati-2hatj+9hatk`

Answer

To find the angle between the given vectors, we will use the formula:

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

where \(\vec{a} \cdot \vec{b}\) is the dot product of the vectors, and \(|\vec{a}|\) and \(|\vec{b}|\) are the magnitudes of the vectors.

### Part (i): Vectors \(\vec{a} = 2\hat{i} - 6\hat{j} + 3\hat{k}\) and \(\vec{b} = \hat{i} + 2\hat{j} - 2\hat{k}\)

1. **Calculate the dot product \(\vec{a} \cdot \vec{b}\)**:
\[ \vec{a} \cdot \vec{b} = (2)(1) + (-6)(2) + (3)(-2) = 2 - 12 - 6 = -16 \]

2. **Calculate the magnitudes \(|\vec{a}|\) and \(|\vec{b}|\)**:
\[ |\vec{a}| = \sqrt{2^2 + (-6)^2 + 3^2} = \sqrt{4 + 36 + 9} = \sqrt{49} = 7 \]
\[ |\vec{b}| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]

3. **Substitute into the cosine formula**:
\[ \cos \theta = \frac{-16}{7 \cdot 3} = \frac{-16}{21} \]

4. **Calculate \(\theta\)**:
\[ \theta = \cos^{-1}\left(\frac{-16}{21}\right) \]

### Part (ii): Vectors \(\vec{a} = 6\hat{i} + 3\hat{j} - 2\hat{k}\) and \(\vec{b} = 4\hat{i} - 2\hat{j} + 9\hat{k}\)

1. **Calculate the dot product \(\vec{a} \cdot \vec{b}\)**:
\[ \vec{a} \cdot \vec{b} = (6)(4) + (3)(-2) + (-2)(9) = 24 - 6 - 18 = 0 \]

2. **Calculate the magnitudes \(|\vec{a}|\) and \(|\vec{b}|\)**:
\[ |\vec{a}| = \sqrt{6^2 + 3^2 + (-2)^2} = \sqrt{36 + 9 + 4} = \sqrt{49} = 7 \]
\[ |\vec{b}| = \sqrt{4^2 + (-2)^2 + 9^2} = \sqrt{16 + 4 + 81} = \sqrt{101} \]

3. **Substitute into the cosine formula**:
\[ \cos \theta = \frac{0}{7 \cdot \sqrt{101}} = 0 \]

4. **Calculate \(\theta\)**:
\[ \theta = \cos^{-1}(0) = 90^\circ \]

### Summary of Results:
- For part (i), \(\theta = \cos^{-1}\left(\frac{-16}{21}\right)\)
- For part (ii), \(\theta = 90^\circ\)