Let A be vector parallel to line of intersection of planes `P_1 and P_2`. Plane `P_1` is parallel to the vectors `2hat(j)+3hat(k) and 4hat(j)-3hat(k)` and that `P_2` is parallel to `hat(j)-hat(k) and 3hat(i)+3hat(j)`, then the angle between vector A and a given vector `2hat(i)+hat(j)-2hat(k)` is

Options

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

Answer

To solve the problem step by step, we will follow the outlined approach in the video transcript.

### Step 1: Find the normal vector of Plane P1
Plane P1 is parallel to the vectors \( \mathbf{v_1} = 2\hat{j} + 3\hat{k} \) and \( \mathbf{v_2} = 4\hat{j} - 3\hat{k} \). We can find the normal vector \( \mathbf{n_1} \) of Plane P1 using the cross product of these two vectors.

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

Calculating the determinant:

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

### Step 2: Find the normal vector of Plane P2
Plane P2 is parallel to the vectors \( \mathbf{u_1} = \hat{j} - \hat{k} \) and \( \mathbf{u_2} = 3\hat{i} + 3\hat{j} \). We find the normal vector \( \mathbf{n_2} \) of Plane P2 using the cross product of these two vectors.

\[ \mathbf{n_2} = \mathbf{u_1} \times \mathbf{u_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & -1 \\ 3 & 3 & 0 \end{vmatrix} \]

Calculating the determinant:

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

### Step 3: Find the direction vector of the line of intersection
The direction vector \( \mathbf{A} \) of the line of intersection of the two planes is given by the cross product of the normals \( \mathbf{n_1} \) and \( \mathbf{n_2} \).

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

Calculating the determinant:

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

### Step 4: Find the angle between vector A and the given vector
The given vector is \( \mathbf{B} = 2\hat{i} + \hat{j} - 2\hat{k} \). To find the angle \( \theta \) between vectors \( \mathbf{A} \) and \( \mathbf{B} \), we use the dot product formula:

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

Calculating \( \mathbf{A} \cdot \mathbf{B} \):

\[ \mathbf{A} \cdot \mathbf{B} = (-54\hat{j} + 54\hat{k}) \cdot (2\hat{i} + \hat{j} - 2\hat{k}) = 0 + (-54) + (54 \cdot -2) = -54 - 108 = -162 \]

Calculating the magnitudes:

\[ |\mathbf{A}| = \sqrt{(-54)^2 + (54)^2} = \sqrt{2916 + 2916} = \sqrt{5832} = 54\sqrt{2} \]
\[ |\mathbf{B}| = \sqrt{(2)^2 + (1)^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \]

Now substituting into the cosine formula:

\[ \cos \theta = \frac{-162}{54\sqrt{2} \cdot 3} = \frac{-162}{162\sqrt{2}} = \frac{-1}{\sqrt{2}} \]

### Step 5: Find the angles
The angles corresponding to \( \cos \theta = -\frac{1}{\sqrt{2}} \) are:

\[ \theta = \frac{3\pi}{4} \quad \text{and} \quad \theta = \frac{5\pi}{4} \]

### Final Answer
The angle between vector \( \mathbf{A} \) and the given vector \( \mathbf{B} \) is \( \frac{3\pi}{4} \) radians or \( 135^\circ \).

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