A tetrahedron has vertices `O(0,0,0),A(1,2,1),B(2,1,3),a n dC(-1,1,2),` then angle between face `O A Ba n dA B C` will be a. `cos^(-1)((17)/(31))` b. `30^0` c. `90^0` d. `cos^(-1)((19)/(35))`

Options

A. `cos^(-1)((17)/(31))`
B. `30^(@)`
C. `90^(@)`
D. `cos^(-1)((19)/(35))`

Explanation

To find the angle between the faces OAB and ABC of the tetrahedron with vertices O(0,0,0), A(1,2,1), B(2,1,3), and C(-1,1,2), we will follow these steps:

### Step 1: Find the position vectors
The position vectors of points O, A, B, and C are:
- \(\vec{O} = (0, 0, 0)\)
- \(\vec{A} = (1, 2, 1)\)
- \(\vec{B} = (2, 1, 3)\)
- \(\vec{C} = (-1, 1, 2)\)

### Step 2: Find vectors OA and OB
The vectors OA and OB can be calculated as follows:
\[ \vec{OA} = \vec{A} - \vec{O} = (1, 2, 1) - (0, 0, 0) = (1, 2, 1) \]
\[ \vec{OB} = \vec{B} - \vec{O} = (2, 1, 3) - (0, 0, 0) = (2, 1, 3) \]

### Step 3: Calculate the normal vector to the plane OAB
The normal vector \( \vec{n_1} \) to the plane OAB can be found using the cross product of vectors OA and OB:
\[ \vec{n_1} = \vec{OA} \times \vec{OB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ 2 & 1 & 3 \end{vmatrix} \]
Calculating this determinant:
\[ \vec{n_1} = \hat{i} \begin{vmatrix} 2 & 1 \\ 1 & 3 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ 2 & 3 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix} \]
\[ = \hat{i} (2 \cdot 3 - 1 \cdot 1) - \hat{j} (1 \cdot 3 - 1 \cdot 2) + \hat{k} (1 \cdot 1 - 2 \cdot 2) \]
\[ = \hat{i} (6 - 1) - \hat{j} (3 - 2) + \hat{k} (1 - 4) \]
\[ = 5\hat{i} - 1\hat{j} - 3\hat{k} \]
Thus, \( \vec{n_1} = (5, -1, -3) \).

### Step 4: Find vectors AB and AC
Next, we find the vectors AB and AC:
\[ \vec{AB} = \vec{B} - \vec{A} = (2, 1, 3) - (1, 2, 1) = (1, -1, 2) \]
\[ \vec{AC} = \vec{C} - \vec{A} = (-1, 1, 2) - (1, 2, 1) = (-2, -1, 1) \]

### Step 5: Calculate the normal vector to the plane ABC
The normal vector \( \vec{n_2} \) to the plane ABC can be found using the cross product of vectors AB and AC:
\[ \vec{n_2} = \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ -2 & -1 & 1 \end{vmatrix} \]
Calculating this determinant:
\[ \vec{n_2} = \hat{i} \begin{vmatrix} -1 & 2 \\ -1 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 2 \\ -2 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & -1 \\ -2 & -1 \end{vmatrix} \]
\[ = \hat{i} (-1 \cdot 1 - 2 \cdot -1) - \hat{j} (1 \cdot 1 - 2 \cdot -2) + \hat{k} (1 \cdot -1 - -1 \cdot -2) \]
\[ = \hat{i} (-1 + 2) - \hat{j} (1 + 4) + \hat{k} (-1 - 2) \]
\[ = 1\hat{i} - 5\hat{j} - 3\hat{k} \]
Thus, \( \vec{n_2} = (1, -5, -3) \).

### Step 6: Find the angle between the two normal vectors
The angle \( \theta \) between the two planes is given by the formula:
\[ \cos \theta = \frac{\vec{n_1} \cdot \vec{n_2}}{|\vec{n_1}| |\vec{n_2}|} \]
Calculating the dot product \( \vec{n_1} \cdot \vec{n_2} \):
\[ \vec{n_1} \cdot \vec{n_2} = (5)(1) + (-1)(-5) + (-3)(-3) = 5 + 5 + 9 = 19 \]

### Step 7: Calculate the magnitudes of the normal vectors
Calculating the magnitudes:
\[ |\vec{n_1}| = \sqrt{5^2 + (-1)^2 + (-3)^2} = \sqrt{25 + 1 + 9} = \sqrt{35} \]
\[ |\vec{n_2}| = \sqrt{1^2 + (-5)^2 + (-3)^2} = \sqrt{1 + 25 + 9} = \sqrt{35} \]

### Step 8: Substitute into the cosine formula
Now substituting back into the cosine formula:
\[ \cos \theta = \frac{19}{\sqrt{35} \cdot \sqrt{35}} = \frac{19}{35} \]

### Step 9: Find the angle
Thus, the angle \( \theta \) is:
\[ \theta = \cos^{-1}\left(\frac{19}{35}\right) \]

### Conclusion
The angle between the faces OAB and ABC is \( \cos^{-1}\left(\frac{19}{35}\right) \).