If `vec(A)=2hat(i)-2hat(j)-hat(k)` and `vec(B)=hat(i)+hat(j)`, then:
(i) Find angle between `vec(A)` and `vec(B)`
(ii) Find the projection of resultant vector of `vec(A)` and `vec(B)` on x-axis.
(iii) Find a vector which is, if added to `vec(A)`, gives a unit vector along y-axis.

Explanation

To solve the given problem step by step, we will break it down into three parts as stated in the question.

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

### (i) Find the angle between \(\vec{A}\) and \(\vec{B}\)

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

\[ \vec{A} \cdot \vec{B} = (2\hat{i} - 2\hat{j} - \hat{k}) \cdot (\hat{i} + \hat{j}) \]
\[ = 2 \cdot 1 + (-2) \cdot 1 + (-1) \cdot 0 = 2 - 2 + 0 = 0 \]

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

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

\[ |\vec{B}| = \sqrt{(1)^2 + (1)^2} = \sqrt{1 + 1} = \sqrt{2} \]

**Step 3: Use the cosine formula to find the angle \(\theta\)**

\[ \cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|} \]
\[ \cos \theta = \frac{0}{3 \cdot \sqrt{2}} = 0 \]

Since \(\cos \theta = 0\), we find that:

\[ \theta = 90^\circ \]

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

**Step 1: Calculate the resultant vector \(\vec{R} = \vec{A} + \vec{B}\)**

\[ \vec{R} = (2\hat{i} - 2\hat{j} - \hat{k}) + (\hat{i} + \hat{j}) = (2 + 1)\hat{i} + (-2 + 1)\hat{j} - \hat{k} \]
\[ = 3\hat{i} - \hat{j} - \hat{k} \]

**Step 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 \(\hat{i}\).

\[ \text{Projection on x-axis} = 3 \]

### (iii) Find a vector which, if added to \(\vec{A}\), gives a unit vector along the y-axis

**Step 1: Let the required vector be \(\vec{C}\)**

We need:
\[ \vec{A} + \vec{C} = \hat{j} \]

**Step 2: Substitute \(\vec{A}\) into the equation**

\[ (2\hat{i} - 2\hat{j} - \hat{k}) + \vec{C} = \hat{j} \]

**Step 3: Rearranging gives us \(\vec{C}\)**

\[ \vec{C} = \hat{j} - (2\hat{i} - 2\hat{j} - \hat{k}) \]
\[ = \hat{j} - 2\hat{i} + 2\hat{j} + \hat{k} \]
\[ = -2\hat{i} + 3\hat{j} + \hat{k} \]

### Summary of Results:
(i) The angle between \(\vec{A}\) and \(\vec{B}\) is \(90^\circ\).
(ii) The projection of the resultant vector on the x-axis is \(3\).
(iii) The vector that, when added to \(\vec{A}\), gives a unit vector along the y-axis is \(-2\hat{i} + 3\hat{j} + \hat{k}\).