1. Key Concepts and Formulas

• Angle Between Two Planes: The angle $\theta$ between two intersecting planes is defined as the acute angle between their normal vectors.
• Formula for Angle: If $\vec{n_1}$ and $\vec{n_2}$ are the normal vectors to two planes, the angle $\theta$ between them is given by: $\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$ The absolute value of the dot product ($|\vec{n_1} \cdot \vec{n_2}|$) ensures that we always find the acute angle, which is the standard convention for the angle between faces of a polyhedron.
• Normal Vector to a Plane (using Cross Product): To find a normal vector to a plane defined by three non-collinear points (e.g., A, B, C), we can take the cross product of two vectors lying in that plane (e.g., $\vec{AB} \times \vec{AC}$).

2. Step-by-Step Solution

Step 1: Identify the Vertices and Faces We are given the vertices of the tetrahedron:

• O(0, 0, 0)
• P(1, 2, 1)
• Q(2, 1, 3)
• R(–1, 1, 2)

We need to find the angle between face OPQ and face PQR.

Step 2: Find the Normal Vector ($\vec{n_1}$) for Face OPQ To find the normal vector for face OPQ, we can use vectors $\vec{OP}$ and $\vec{OQ}$, which lie in the plane of the face.

• Vector $\vec{OP} = P - O = \langle 1-0, 2-0, 1-0 \rangle = \langle 1, 2, 1 \rangle$.
• Vector $\vec{OQ} = Q - O = \langle 2-0, 1-0, 3-0 \rangle = \langle 2, 1, 3 \rangle$.

Now, we calculate their cross product to get the normal vector $\vec{n_1}$: $\vec{n_1} = \vec{OP} \times \vec{OQ} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 1 \\ 2 & 1 & 3 \end{vmatrix}$ $\vec{n_1} = \mathbf{i}((2)(3) - (1)(1)) - \mathbf{j}((1)(3) - (1)(2)) + \mathbf{k}((1)(1) - (2)(2))$ $\vec{n_1} = \mathbf{i}(6 - 1) - \mathbf{j}(3 - 2) + \mathbf{k}(1 - 4)$ $\vec{n_1} = 5\mathbf{i} - 1\mathbf{j} - 3\mathbf{k} = \langle 5, -1, -3 \rangle$

Step 3: Calculate the Magnitude of $\vec{n_1}$ The magnitude of $\vec{n_1}$ is: $|\vec{n_1}| = \sqrt{5^2 + (-1)^2 + (-3)^2} = \sqrt{25 + 1 + 9} = \sqrt{35}$

Step 4: Find the Normal Vector ($\vec{n_2}$) for Face PQR To find the normal vector for face PQR, we use two vectors lying in that plane, originating from a common vertex. Let's use $\vec{PQ}$ and $\vec{PR}$.

• Vector $\vec{PQ} = Q - P = \langle 2-1, 1-2, 3-1 \rangle = \langle 1, -1, 2 \rangle$.
• Vector $\vec{PR} = R - P = \langle -1-1, 1-2, 2-1 \rangle = \langle -2, -1, 1 \rangle$.

Now, we calculate their cross product to get the normal vector $\vec{n_2}$: $\vec{n_2} = \vec{PQ} \times \vec{PR} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 2 \\ -2 & -1 & 1 \end{vmatrix}$ $\vec{n_2} = \mathbf{i}((-1)(1) - (2)(-1)) - \mathbf{j}((1)(1) - (2)(-2)) + \mathbf{k}((1)(-1) - (-1)(-2))$ $\vec{n_2} = \mathbf{i}(-1 + 2) - \mathbf{j}(1 + 4) + \mathbf{k}(-1 - 2)$ $\vec{n_2} = 1\mathbf{i} - 5\mathbf{j} - 3\mathbf{k} = \langle 1, -5, -3 \rangle$

Step 5: Calculate the Magnitude of $\vec{n_2}$ The magnitude of $\vec{n_2}$ is: $|\vec{n_2}| = \sqrt{1^2 + (-5)^2 + (-3)^2} = \sqrt{1 + 25 + 9} = \sqrt{35}$

Step 6: Calculate the Dot Product of $\vec{n_1}$ and $\vec{n_2}$ Now, we find the dot product of the two normal vectors: $\vec{n_1} \cdot \vec{n_2} = (5)(1) + (-1)(-5) + (-3)(-3)$ $\vec{n_1} \cdot \vec{n_2} = 5 + 5 + 9 = 19$

Step 7: Apply the Angle Formula Substitute the calculated values into the formula for the angle $\theta$ between the planes: $\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$ $\cos \theta = \frac{|19|}{\sqrt{35} \cdot \sqrt{35}}$ $\cos \theta = \frac{19}{35}$ Therefore, the angle $\theta$ between the faces OPQ and PQR is: $\theta = \cos^{-1}\left(\frac{19}{35}\right)$

3. Common Mistakes & Tips

• Careful with Cross Product Signs: The calculation of the cross product involves determinants, and sign errors (especially with the $-\mathbf{j}$ term) are common. Double-check each component calculation.
• Consistent Vector Subtraction: When forming vectors from points (e.g., $\vec{PQ} = Q - P$), ensure you subtract the coordinates in the correct order.
• Absolute Value for Acute Angle: Always remember to take the absolute value of the dot product in the numerator ($|\vec{n_1} \cdot \vec{n_2}|$) to ensure you find the acute angle between the planes, as per the standard definition.

4. Summary

To find the angle between two faces of a tetrahedron, we first determine the normal vector for each face. This is achieved by taking the cross product of two vectors lying within each respective plane. For face OPQ, we used $\vec{OP}$ and $\vec{OQ}$ to find $\vec{n_1} = \langle 5, -1, -3 \rangle$. For face PQR, we used $\vec{PQ}$ and $\vec{PR}$ to find $\vec{n_2} = \langle 1, -5, -3 \rangle$. We then calculated the magnitudes of these normal vectors, both found to be $\sqrt{35}$. Finally, we used the dot product formula for the angle between two vectors, which resulted in $\cos \theta = \frac{19}{35}$.

The final answer is $\boxed{\text{cos}^{-1} \left( \frac{19}{35} \right)}$, which corresponds to option (C).