Let `vecA` be a vector parallel to the line of intersection of the planes `P_(1) and P_(2)`. The plane `P_(1)` is parallel to vectors `2hatj+3hatk and 4hatj-3hatk` while plane `P_(2)` is parallel to the vectors `hatj-hatk` and `hati+hatj`. The acute angle between `vecA` and `2hati+hatj-2hatk` is

Options

A. `(pi)/(6)`
B. `(pi)/(4)`
C. `(pi)/(3)`
D. `(5pi)/(12)`

Answer

To solve the problem, we need to find the acute angle between the vector \(\vec{A}\) (which is parallel to the line of intersection of the planes \(P_1\) and \(P_2\)) and the vector \(2\hat{i} + \hat{j} - 2\hat{k}\).

### Step 1: Find the normal vector of plane \(P_1\)

The plane \(P_1\) is parallel to the vectors \(2\hat{j} + 3\hat{k}\) and \(4\hat{j} - 3\hat{k}\). The normal vector \(\vec{n_1}\) of the plane can be found using the cross product of these two vectors.

\[ \vec{n_1} = (2\hat{j} + 3\hat{k}) \times (4\hat{j} - 3\hat{k}) \]

Calculating the cross product:

\[ \vec{n_1} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 2 & 3 \\ 0 & 4 & -3 \end{vmatrix} \]

Calculating the determinant:

\[ = \hat{i}(2 \cdot (-3) - 3 \cdot 4) - \hat{j}(0 \cdot (-3) - 3 \cdot 0) + \hat{k}(0 \cdot 4 - 0 \cdot 2) \]
\[ = \hat{i}(-6 - 12) + 0 + 0 = -18\hat{i} \]

Thus, \(\vec{n_1} = -18\hat{i}\).

### Step 2: Find the normal vector of plane \(P_2\)

The plane \(P_2\) is parallel to the vectors \(\hat{j} - \hat{k}\) and \(\hat{i} + \hat{j}\). The normal vector \(\vec{n_2}\) of the plane can be found using the cross product of these two vectors.

\[ \vec{n_2} = (\hat{j} - \hat{k}) \times (\hat{i} + \hat{j}) \]

Calculating the cross product:

\[ \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & -1 \\ 1 & 1 & 0 \end{vmatrix} \]

Calculating the determinant:

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

Thus, \(\vec{n_2} = \hat{i} - \hat{j} + \hat{k}\).

### Step 3: Find the direction vector \(\vec{A}\)

The direction vector \(\vec{A}\) that is parallel to the line of intersection of the planes \(P_1\) and \(P_2\) can be found using the cross product of the normal vectors \(\vec{n_1}\) and \(\vec{n_2}\).

\[ \vec{A} = \vec{n_1} \times \vec{n_2} = (-18\hat{i}) \times (\hat{i} - \hat{j} + \hat{k}) \]

Calculating the cross product:

\[ \vec{A} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -18 & 0 & 0 \\ 1 & -1 & 1 \end{vmatrix} \]

Calculating the determinant:

\[ = \hat{i}(0 \cdot 1 - 0 \cdot (-1)) - \hat{j}(-18 \cdot 1 - 0 \cdot 0) + \hat{k}(-18 \cdot (-1) - 0 \cdot 1) \]
\[ = 0\hat{i} + 18\hat{j} + 18\hat{k} \]

Thus, \(\vec{A} = 0\hat{i} + 18\hat{j} + 18\hat{k}\).

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

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

To find the angle \(\theta\) between \(\vec{A}\) and \(\vec{B}\), we use the formula:

\[ \cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|} \]

Calculating the dot product \(\vec{A} \cdot \vec{B}\):

\[ \vec{A} \cdot \vec{B} = (0)(2) + (18)(1) + (18)(-2) = 0 + 18 - 36 = -18 \]

Calculating the magnitudes:

\[ |\vec{A}| = \sqrt{0^2 + 18^2 + 18^2} = \sqrt{0 + 324 + 324} = \sqrt{648} = 18\sqrt{2} \]

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

Now substituting into the cosine formula:

\[ \cos \theta = \frac{-18}{(18\sqrt{2})(3)} = \frac{-18}{54\sqrt{2}} = \frac{-1}{3\sqrt{2}} \]

### Step 5: Finding the angle \(\theta\)

To find \(\theta\):

\[ \theta = \cos^{-1}\left(\frac{-1}{3\sqrt{2}}\right) \]

Since this angle is obtuse, the acute angle \(\theta'\) can be found as:

\[ \theta' = \pi - \theta \]

### Final Answer

The acute angle between \(\vec{A}\) and \(2\hat{i} + \hat{j} - 2\hat{k}\) is:

\[ \theta' = \frac{\pi}{4} \]