Let `vec a ` be vector parallel to line of intersection of planes `P_1 and P_2` through origin. If `P_1`is parallel to the vectors `2 bar j + 3 bar k and 4 bar j - 3 bar k` and `P_2` is parallel to `bar j - bar k` and ` 3 bar I + 3 bar j `, then the angle between `vec a` and `2 bar i +bar j - 2 bar k` is :

Options

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

Explanation

To solve the problem, we will follow these steps:

### Step 1: Find the normal vector of plane P1
Plane P1 is parallel to the vectors \( \vec{v_1} = 2\hat{j} + 3\hat{k} \) and \( \vec{v_2} = 4\hat{j} - 3\hat{k} \). The normal vector \( \vec{n_1} \) of plane P1 can be found using the cross product of \( \vec{v_1} \) and \( \vec{v_2} \).

\[ \vec{n_1} = \vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 2 & 4 \\ 0 & 3 & -3 \end{vmatrix} \]

Calculating the determinant:

\[ \vec{n_1} = \hat{i}(2 \cdot (-3) - 4 \cdot 3) - \hat{j}(0 \cdot (-3) - 0 \cdot 4) + \hat{k}(0 \cdot 3 - 0 \cdot 2) \]
\[ = \hat{i}(-6 - 12) = -18\hat{i} \]

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

### Step 2: Find the normal vector of plane P2
Plane P2 is parallel to the vectors \( \vec{w_1} = \hat{j} - \hat{k} \) and \( \vec{w_2} = 3\hat{i} + 3\hat{j} \). The normal vector \( \vec{n_2} \) of plane P2 can be found using the cross product of \( \vec{w_1} \) and \( \vec{w_2} \).

\[ \vec{n_2} = \vec{w_1} \times \vec{w_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & -1 \\ 3 & 3 & 0 \end{vmatrix} \]

Calculating the determinant:

\[ \vec{n_2} = \hat{i}(1 \cdot 0 - (-1) \cdot 3) - \hat{j}(0 \cdot 0 - (-1) \cdot 3) + \hat{k}(0 \cdot 3 - 1 \cdot 3) \]
\[ = \hat{i}(0 + 3) - \hat{j}(0 + 3) - 3\hat{k} \]
\[ = 3\hat{i} - 3\hat{j} - 3\hat{k} \]

So, \( \vec{n_2} = 3\hat{i} - 3\hat{j} - 3\hat{k} \).

### Step 3: Find the direction vector of the line of intersection
The direction vector \( \vec{a} \) of the line of intersection of the two planes is given by the cross product of their normal vectors \( \vec{n_1} \) and \( \vec{n_2} \).

\[ \vec{a} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -18 & 0 & 0 \\ 3 & -3 & -3 \end{vmatrix} \]

Calculating the determinant:

\[ \vec{a} = \hat{i}(0 \cdot (-3) - 0 \cdot (-3)) - \hat{j}(-18 \cdot (-3) - 0 \cdot 3) + \hat{k}(-18 \cdot (-3) - 0 \cdot 0) \]
\[ = 0\hat{i} - \hat{j}(54) + \hat{k}(54) \]
\[ = 0\hat{i} - 54\hat{j} + 54\hat{k} \]

So, \( \vec{a} = 54(\hat{k} - \hat{j}) \).

### Step 4: Find the angle between vector \( \vec{a} \) and vector \( \vec{b} = 2\hat{i} + \hat{j} - 2\hat{k} \)
To find the angle \( \theta \) between \( \vec{a} \) and \( \vec{b} \), we use the dot product formula:

\[ \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} \]

Calculating \( \vec{a} \cdot \vec{b} \):

\[ \vec{a} \cdot \vec{b} = 54(\hat{k} - \hat{j}) \cdot (2\hat{i} + \hat{j} - 2\hat{k}) = 54(0 + (-54) + (-108)) = 54(0 + 1 + 2) = 54 \cdot 1 = 54 \]

Calculating \( |\vec{a}| \):

\[ |\vec{a}| = |54(\hat{k} - \hat{j})| = 54\sqrt{1^2 + (-1)^2} = 54\sqrt{2} \]

Calculating \( |\vec{b}| \):

\[ |\vec{b}| = |2\hat{i} + \hat{j} - 2\hat{k}| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \]

Now substituting back into the cosine formula:

\[ \cos \theta = \frac{54}{54\sqrt{2} \cdot 3} = \frac{1}{\sqrt{2}} \]

Thus,

\[ \theta = \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) = 45^\circ \]

### Final Answer
The angle between \( \vec{a} \) and \( 2\hat{i} + \hat{j} - 2\hat{k} \) is \( 45^\circ \).

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