A non zero vector `veca` is parallel to the line of intersection of the plane determined by the vectors `hati,hati+hatj`and the plane determined by the vectors `hati-hatj , hati + hatk`. The angle between `veca` and `hati-2hatj + 2hatk` can be

Answer

To solve the problem step by step, we will find the vector that is parallel to the line of intersection of the two planes defined by the given vectors and then determine the angle between this vector and the specified vector.

### Step 1: Identify the normal vectors of the planes

The first plane is determined by the vectors \(\hat{i}\) and \(\hat{i} + \hat{j}\). We can find the normal vector \(n_1\) of this plane using the cross product of these two vectors.

1. **Vectors**:
- \(\vec{v_1} = \hat{i}\) (which can be represented as \((1, 0, 0)\))
- \(\vec{v_2} = \hat{i} + \hat{j}\) (which can be represented as \((1, 1, 0)\))

2. **Cross Product**:
\[ n_1 = \vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & 0 \\ 1 & 1 & 0 \end{vmatrix} = (0 - 0)\hat{i} - (0 - 0)\hat{j} + (1 - 0)\hat{k} = \hat{k} \]
Thus, \(n_1 = \hat{k}\).

### Step 2: Find the normal vector of the second plane

The second plane is determined by the vectors \(\hat{i} - \hat{j}\) and \(\hat{i} + \hat{k}\).

1. **Vectors**:
- \(\vec{v_3} = \hat{i} - \hat{j}\) (which can be represented as \((1, -1, 0)\))
- \(\vec{v_4} = \hat{i} + \hat{k}\) (which can be represented as \((1, 0, 1)\))

2. **Cross Product**:
\[ n_2 = \vec{v_3} \times \vec{v_4} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 0 \\ 1 & 0 & 1 \end{vmatrix} = (-1 - 0)\hat{i} - (0 - 1)\hat{j} + (0 - (-1))\hat{k} = -\hat{i} + \hat{j} + \hat{k} \]
Thus, \(n_2 = -\hat{i} + \hat{j} + \hat{k}\).

### Step 3: Find the line of intersection of the two planes

The direction vector of the line of intersection of the two planes can be found using the cross product of the normal vectors \(n_1\) and \(n_2\).

\[ \vec{a} = n_1 \times n_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 0 & 1 \\ -1 & 1 & 1 \end{vmatrix} = (0 - 1)\hat{i} - (0 - (-1))\hat{j} + (0 - 0)\hat{k} = -\hat{i} + \hat{j} \]

Thus, \(\vec{a} = -\hat{i} + \hat{j}\).

### Step 4: Find the angle between \(\vec{a}\) and the vector \(\hat{i} - 2\hat{j} + 2\hat{k}\)

Let \(\vec{b} = \hat{i} - 2\hat{j} + 2\hat{k}\).

1. **Dot Product**:
\[ \vec{a} \cdot \vec{b} = (-1)(1) + (1)(-2) + (0)(2) = -1 - 2 + 0 = -3 \]

2. **Magnitudes**:
\[ |\vec{a}| = \sqrt{(-1)^2 + (1)^2 + 0^2} = \sqrt{1 + 1} = \sqrt{2} \]
\[ |\vec{b}| = \sqrt{(1)^2 + (-2)^2 + (2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]

3. **Cosine of the angle**:
\[ \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{-3}{\sqrt{2} \cdot 3} = \frac{-1}{\sqrt{2}} \]

4. **Finding the angle**:
\[ \theta = \cos^{-1}\left(-\frac{1}{\sqrt{2}}\right) = \frac{3\pi}{4} \text{ radians or } 135^\circ \]

### Final Answer:
The angle between the vector \(\vec{a}\) and \(\hat{i} - 2\hat{j} + 2\hat{k}\) can be \(135^\circ\) or \(\frac{3\pi}{4}\) radians.

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