If `vecA= 2hati-2hatj-hatk and vecB= hati+hatj`, then :
(a) Find angle between `vecA and vecB`.
(b) Find the projection of resultant vector of `vecA and vecB` on x-axis.

Explanation

To solve the given problem step by step, we will address both parts of the question: finding the angle between the vectors and finding the projection of the resultant vector on the x-axis.

### Given:
- \(\vec{A} = 2\hat{i} - 2\hat{j} - \hat{k}\)
- \(\vec{B} = \hat{i} + \hat{j}\)

### Part (a): Find the angle between \(\vec{A}\) and \(\vec{B}\).

1. **Calculate the dot product of \(\vec{A}\) and \(\vec{B}\)**:
\[ \vec{A} \cdot \vec{B} = (2\hat{i} - 2\hat{j} - \hat{k}) \cdot (\hat{i} + \hat{j}) \]
\[ = 2 \cdot 1 + (-2) \cdot 1 + (-1) \cdot 0 = 2 - 2 + 0 = 0 \]

2. **Calculate the magnitudes of \(\vec{A}\) and \(\vec{B}\)**:
\[ |\vec{A}| = \sqrt{(2)^2 + (-2)^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \]
\[ |\vec{B}| = \sqrt{(1)^2 + (1)^2} = \sqrt{1 + 1} = \sqrt{2} \]

3. **Use the dot product to find the cosine of the angle**:
\[ \cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|} = \frac{0}{3 \cdot \sqrt{2}} = 0 \]

4. **Determine the angle \(\theta\)**:
Since \(\cos \theta = 0\), we find:
\[ \theta = 90^\circ \]

### Part (b): Find the projection of the resultant vector of \(\vec{A}\) and \(\vec{B}\) on the x-axis.

1. **Calculate the resultant vector \(\vec{R} = \vec{A} + \vec{B}\)**:
\[ \vec{R} = (2\hat{i} - 2\hat{j} - \hat{k}) + (\hat{i} + \hat{j}) \]
\[ = (2 + 1)\hat{i} + (-2 + 1)\hat{j} + (-1)\hat{k} = 3\hat{i} - \hat{j} - \hat{k} \]

2. **Find the projection of \(\vec{R}\) on the x-axis**:
The projection of a vector on the x-axis is simply the coefficient of \(\hat{i}\) in the vector.
\[ \text{Projection of } \vec{R} \text{ on x-axis} = 3 \]

### Final Answers:
(a) The angle between \(\vec{A}\) and \(\vec{B}\) is \(90^\circ\).
(b) The projection of the resultant vector on the x-axis is \(3\).