A non-zero vector `bara` is parallel to the line of intersection of the plane `P_1`determined by `hati+hatj and hati-2hatj` and plane `P_2` determined by vector `2hati+hatj and 3hati+2hatk` then angle between `bara` and vector `hati-2hatj+2hatk` is

Options

A. `pi/4`
B. `pi/2`
C. `pi/3`
D. None of these

Answer

To solve the problem, we need to find the angle between the vector \( \bar{a} \) (which is parallel to the line of intersection of two planes) and the vector \( \bar{b} = \hat{i} - 2\hat{j} + 2\hat{k} \).

### Step 1: Determine the normals of the planes \( P_1 \) and \( P_2 \)

**Plane \( P_1 \)** is determined by the vectors \( \hat{i} + \hat{j} \) and \( \hat{i} - 2\hat{j} \).

To find the normal vector \( \bar{n_1} \) of plane \( P_1 \), we calculate the cross product of these two vectors:

\[ \bar{n_1} = (\hat{i} + \hat{j}) \times (\hat{i} - 2\hat{j}) \]

Calculating the cross product:

\[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 1 & -2 & 0 \end{vmatrix} \]

Expanding this determinant:

\[ \bar{n_1} = \hat{i}(1 \cdot 0 - (-2) \cdot 0) - \hat{j}(1 \cdot 0 - 1 \cdot 0) + \hat{k}(1 \cdot (-2) - 1 \cdot 1) \]

This simplifies to:

\[ \bar{n_1} = \hat{k}(-2 - 1) = -3\hat{k} \]

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

**Plane \( P_2 \)** is determined by the vectors \( 2\hat{i} + \hat{j} \) and \( 3\hat{i} + 2\hat{k} \).

To find the normal vector \( \bar{n_2} \) of plane \( P_2 \), we calculate the cross product of these two vectors:

\[ \bar{n_2} = (2\hat{i} + \hat{j}) \times (3\hat{i} + 2\hat{k}) \]

Calculating the cross product:

\[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 0 \\ 3 & 0 & 2 \end{vmatrix} \]

Expanding this determinant:

\[ \bar{n_2} = \hat{i}(1 \cdot 2 - 0 \cdot 0) - \hat{j}(2 \cdot 2 - 0 \cdot 3) + \hat{k}(2 \cdot 0 - 1 \cdot 3) \]

This simplifies to:

\[ \bar{n_2} = 2\hat{i} - 4\hat{j} - 3\hat{k} \]

### Step 3: Find the direction vector \( \bar{a} \)

The direction vector \( \bar{a} \) of the line of intersection of the two planes is given by the cross product of the normals \( \bar{n_1} \) and \( \bar{n_2} \):

\[ \bar{a} = \bar{n_1} \times \bar{n_2} = (-3\hat{k}) \times (2\hat{i} - 4\hat{j} - 3\hat{k}) \]

Calculating this cross product:

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

Expanding this determinant:

\[ \bar{a} = \hat{i}(0 \cdot -3 - (-3) \cdot -4) - \hat{j}(0 \cdot -3 - (-3) \cdot 2) + \hat{k}(0 \cdot -4 - 0 \cdot 2) \]

This simplifies to:

\[ \bar{a} = \hat{i}(0 - 12) - \hat{j}(0 - 6) + 0 = -12\hat{i} + 6\hat{j} \]

### Step 4: Find the angle between \( \bar{a} \) and \( \bar{b} \)

Now we need to find the angle \( \theta \) between \( \bar{a} = -12\hat{i} + 6\hat{j} \) and \( \bar{b} = \hat{i} - 2\hat{j} + 2\hat{k} \).

Using the formula for the cosine of the angle between two vectors:

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

Calculating the dot product \( \bar{a} \cdot \bar{b} \):

\[ \bar{a} \cdot \bar{b} = (-12)(1) + (6)(-2) + (0)(2) = -12 - 12 + 0 = -24 \]

Calculating the magnitudes:

\[ |\bar{a}| = \sqrt{(-12)^2 + 6^2} = \sqrt{144 + 36} = \sqrt{180} = 6\sqrt{5} \]

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

Now substituting back into the cosine formula:

\[ \cos \theta = \frac{-24}{(6\sqrt{5})(3)} = \frac{-24}{18\sqrt{5}} = \frac{-4}{3\sqrt{5}} \]

Since \( \cos \theta \) is negative, the angle \( \theta \) is obtuse.

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

To find \( \theta \):

\[ \theta = \cos^{-1}\left(\frac{-4}{3\sqrt{5}}\right) \]

Since the cosine is negative, we can conclude that:

\[ \theta = 90^\circ \text{ or } \frac{\pi}{2} \text{ radians} \]

### Final Answer

The angle between \( \bar{a} \) and \( \bar{b} \) is \( \frac{\pi}{2} \) radians or \( 90^\circ \).

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