A non-zero vector is a parallel to theline of intersection of the plane determined by the vectors `i,i+j` and the plane dtermined by the vectors `i-j,i+k`. The angle between a and the vector `i-2j+2k` is ........

Answer

To solve the problem, we need to find a non-zero vector that is parallel to the line of intersection of two planes defined by the given vectors. We will then find the angle between this vector and the vector \( \mathbf{d} = \mathbf{i} - 2\mathbf{j} + 2\mathbf{k} \).

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

1. **Plane 1** is determined by the vectors \( \mathbf{a_1} = \mathbf{i} \) and \( \mathbf{a_2} = \mathbf{i} + \mathbf{j} \).
- The normal vector \( \mathbf{n_1} \) can be found using the cross product:
\[ \mathbf{n_1} = \mathbf{a_1} \times \mathbf{a_2} = \mathbf{i} \times (\mathbf{i} + \mathbf{j}) = \mathbf{i} \times \mathbf{i} + \mathbf{i} \times \mathbf{j} = \mathbf{0} + \mathbf{k} = \mathbf{k} \]

2. **Plane 2** is determined by the vectors \( \mathbf{b_1} = \mathbf{i} - \mathbf{j} \) and \( \mathbf{b_2} = \mathbf{i} + \mathbf{k} \).
- The normal vector \( \mathbf{n_2} \) can be found using the cross product:
\[ \mathbf{n_2} = \mathbf{b_1} \times \mathbf{b_2} = (\mathbf{i} - \mathbf{j}) \times (\mathbf{i} + \mathbf{k}) \]
- Expanding this:
\[ \mathbf{n_2} = \mathbf{i} \times \mathbf{i} + \mathbf{i} \times \mathbf{k} - \mathbf{j} \times \mathbf{i} - \mathbf{j} \times \mathbf{k} = \mathbf{0} + \mathbf{j} + \mathbf{j} + \mathbf{i} = \mathbf{i} + 2\mathbf{j} \]

### Step 2: Find the direction vector of the line of intersection

The direction vector \( \mathbf{d} \) of the line of intersection of the two planes can be found using the cross product of the normal vectors:
\[ \mathbf{d} = \mathbf{n_1} \times \mathbf{n_2} = \mathbf{k} \times (\mathbf{i} + 2\mathbf{j}) \]
Calculating this:
\[ \mathbf{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 0 & 1 \\ 1 & 2 & 0 \end{vmatrix} = \mathbf{i}(0 - 2) - \mathbf{j}(0 - 1) + \mathbf{k}(0 - 0) = -2\mathbf{i} + \mathbf{j} \]
Thus, the direction vector of the line of intersection is:
\[ \mathbf{d} = -2\mathbf{i} + \mathbf{j} \]

### Step 3: Find the angle between the vector \( \mathbf{d} \) and \( \mathbf{v} = \mathbf{i} - 2\mathbf{j} + 2\mathbf{k} \)

To find the angle \( \theta \) between the vectors \( \mathbf{d} \) and \( \mathbf{v} \), we use the dot product formula:
\[ \cos \theta = \frac{\mathbf{d} \cdot \mathbf{v}}{|\mathbf{d}| |\mathbf{v}|} \]

1. **Calculate the dot product \( \mathbf{d} \cdot \mathbf{v} \)**:
\[ \mathbf{d} \cdot \mathbf{v} = (-2\mathbf{i} + \mathbf{j}) \cdot (\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}) = -2(1) + 1(-2) + 0(2) = -2 - 2 = -4 \]

2. **Calculate the magnitudes \( |\mathbf{d}| \) and \( |\mathbf{v}| \)**:
- For \( \mathbf{d} \):
\[ |\mathbf{d}| = \sqrt{(-2)^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5} \]
- For \( \mathbf{v} \):
\[ |\mathbf{v}| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]

3. **Substituting into the cosine formula**:
\[ \cos \theta = \frac{-4}{\sqrt{5} \cdot 3} = \frac{-4}{3\sqrt{5}} \]

4. **Finding \( \theta \)**:
To find the angle \( \theta \):
\[ \theta = \cos^{-1}\left(\frac{-4}{3\sqrt{5}}\right) \]

### Final Answer
The angle between the vector \( \mathbf{a} \) (the direction of the line of intersection) and the vector \( \mathbf{d} \) is:
\[ \theta = \cos^{-1}\left(\frac{-4}{3\sqrt{5}}\right) \]