A plane is parallel to the vectors `hati+hatj+hatk` and `2hatk` and another plane is parallel to the vectors `hati+hatj` and `hati-hatk`. The acute angle between the line of intersection of the two planes and the vector `hati-hatj+hatk` is

Options

A. `(pi)/6`
B. `(pi)/4`
C. `(pi)/3`
D. `(pi)/2`

Explanation

To solve the problem, we need to find the acute angle between the line of intersection of two planes and a given vector. Here are the steps to find the solution:

### Step 1: Identify the normal vectors of the planes
The first plane is parallel to the vectors **i + j + k** and **2k**. Therefore, we can find the normal vector \( \mathbf{n_1} \) of the first plane by taking the cross product of these two vectors.

\[ \mathbf{n_1} = (1, 1, 1) \times (0, 0, 2) \]

### Step 2: Calculate the cross product for the first plane
Using the determinant to compute the cross product:

\[ \mathbf{n_1} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 1 \\ 0 & 0 & 2 \end{vmatrix} = \mathbf{i}(1 \cdot 2 - 1 \cdot 0) - \mathbf{j}(1 \cdot 2 - 1 \cdot 0) + \mathbf{k}(1 \cdot 0 - 1 \cdot 0) \]
\[ = 2\mathbf{i} - 2\mathbf{j} + 0\mathbf{k} = 2\mathbf{i} - 2\mathbf{j} \]

### Step 3: Identify the normal vector for the second plane
The second plane is parallel to the vectors **i + j** and **i - k**. We can find the normal vector \( \mathbf{n_2} \) of the second plane similarly.

\[ \mathbf{n_2} = (1, 1, 0) \times (1, 0, -1) \]

### Step 4: Calculate the cross product for the second plane
Using the determinant to compute the cross product:

\[ \mathbf{n_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 0 \\ 1 & 0 & -1 \end{vmatrix} = \mathbf{i}(1 \cdot -1 - 0 \cdot 0) - \mathbf{j}(1 \cdot -1 - 0 \cdot 1) + \mathbf{k}(1 \cdot 0 - 1 \cdot 1) \]
\[ = -\mathbf{i} + \mathbf{j} - \mathbf{k} \]

### Step 5: Find the direction vector of the line of intersection
The direction vector \( \mathbf{L} \) of the line of intersection of the two planes is given by the cross product of the normal vectors \( \mathbf{n_1} \) and \( \mathbf{n_2} \).

\[ \mathbf{L} = \mathbf{n_1} \times \mathbf{n_2} = (2, -2, 0) \times (-1, 1, -1) \]

### Step 6: Calculate the cross product for the direction vector
Using the determinant to compute the cross product:

\[ \mathbf{L} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -2 & 0 \\ -1 & 1 & -1 \end{vmatrix} = \mathbf{i}((-2)(-1) - (0)(1)) - \mathbf{j}((2)(-1) - (0)(-1)) + \mathbf{k}((2)(1) - (-2)(-1)) \]
\[ = 2\mathbf{i} + 2\mathbf{j} + 0\mathbf{k} = 2\mathbf{i} + 2\mathbf{j} \]

### Step 7: Find the angle between the line of intersection and the vector
We need to find the angle \( \theta \) between the vector \( \mathbf{v} = \mathbf{i} - \mathbf{j} + \mathbf{k} \) and the direction vector \( \mathbf{L} = 2\mathbf{i} + 2\mathbf{j} \).

Using the dot product formula:

\[ \cos \theta = \frac{\mathbf{L} \cdot \mathbf{v}}{|\mathbf{L}| |\mathbf{v}|} \]

### Step 8: Calculate the dot product and magnitudes
1. Calculate \( \mathbf{L} \cdot \mathbf{v} \):
\[ \mathbf{L} \cdot \mathbf{v} = (2)(1) + (2)(-1) + (0)(1) = 2 - 2 + 0 = 0 \]

2. Calculate the magnitudes:
\[ |\mathbf{L}| = \sqrt{(2^2 + 2^2 + 0^2)} = \sqrt{8} = 2\sqrt{2} \]
\[ |\mathbf{v}| = \sqrt{(1^2 + (-1)^2 + 1^2)} = \sqrt{3} \]

### Step 9: Substitute into the cosine formula
Since \( \mathbf{L} \cdot \mathbf{v} = 0 \):
\[ \cos \theta = \frac{0}{(2\sqrt{2})(\sqrt{3})} = 0 \]

### Step 10: Find the angle
Thus, \( \theta = \frac{\pi}{2} \) radians or 90 degrees.

### Final Answer
The acute angle between the line of intersection of the two planes and the vector \( \mathbf{i} - \mathbf{j} + \mathbf{k} \) is \( \frac{\pi}{2} \) radians.

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