The angle between `hati` and line of the intersection of the plane `vecr.(hati+2hatj+3hatk)=0andvecr.(3hati+3hatj+hatk)=0` is

Options

A. `cos^(-1)((1)/(3))`
B. `cos^(-1)((1)/(sqrt3))`
C. `cos^(-1)((2)/(sqrt3))`
D. none of these

Explanation

To find the angle between the vector \( \hat{i} \) and the line of intersection of the two planes given by the equations \( \vec{r} \cdot (\hat{i} + 2\hat{j} + 3\hat{k}) = 0 \) and \( \vec{r} \cdot (3\hat{i} + 3\hat{j} + \hat{k}) = 0 \), we will follow these steps:

### Step 1: Identify the normal vectors of the planes
The normal vector of the first plane is \( \vec{n_1} = \hat{i} + 2\hat{j} + 3\hat{k} \) and for the second plane, it is \( \vec{n_2} = 3\hat{i} + 3\hat{j} + \hat{k} \).

### Step 2: Compute the cross product of the normal vectors
The direction vector of the line of intersection of the two planes can be found using the cross product of the normal vectors \( \vec{n_1} \) and \( \vec{n_2} \).

\[ \vec{d} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 3 & 3 & 1 \end{vmatrix} \]

Calculating the determinant:

\[ \vec{d} = \hat{i} \left( 2 \cdot 1 - 3 \cdot 3 \right) - \hat{j} \left( 1 \cdot 1 - 3 \cdot 3 \right) + \hat{k} \left( 1 \cdot 3 - 2 \cdot 3 \right) \]

\[ = \hat{i} (2 - 9) - \hat{j} (1 - 9) + \hat{k} (3 - 6) \]

\[ = -7\hat{i} + 8\hat{j} - 3\hat{k} \]

### Step 3: Find the angle between \( \hat{i} \) and the direction vector \( \vec{d} \)
To find the angle \( \theta \) between \( \hat{i} \) and \( \vec{d} \), we use the formula:

\[ \cos \theta = \frac{\vec{i} \cdot \vec{d}}{|\vec{i}| |\vec{d}|} \]

Where \( \vec{i} \cdot \vec{d} = 1 \cdot (-7) + 0 \cdot 8 + 0 \cdot (-3) = -7 \).

### Step 4: Calculate the magnitudes
The magnitude of \( \hat{i} \) is \( 1 \).

The magnitude of \( \vec{d} \) is:

\[ |\vec{d}| = \sqrt{(-7)^2 + 8^2 + (-3)^2} = \sqrt{49 + 64 + 9} = \sqrt{122} \]

### Step 5: Substitute into the cosine formula
Now substituting back into the cosine formula:

\[ \cos \theta = \frac{-7}{1 \cdot \sqrt{122}} = \frac{-7}{\sqrt{122}} \]

### Step 6: Find the angle \( \theta \)
To find \( \theta \):

\[ \theta = \cos^{-1}\left(\frac{-7}{\sqrt{122}}\right) \]

### Final Answer
Thus, the angle between \( \hat{i} \) and the line of intersection of the two planes is:

\[ \theta = \cos^{-1}\left(\frac{-7}{\sqrt{122}}\right) \]