Animated Solution for Mathematics - Three Dimensional Geometry: A tetrahedron has vertices $P(1,2,1)$, $Q(2,1,3)$, $R(-1,1,2)$ and $O(0,0,0)$. The angle between the faces $OPQ$ and $PQR$ is :

Imagine you are standing in a 3D coordinate system. You have four points: $O(0,0,0)$, $P(1,2,1)$, $Q(2,1,3)$, and $R(-1,1,2)$. These points form a tetrahedron, a beautiful, sharp-edged pyramid.

Our goal is to find the angle between two of its faces: $OPQ$ and $PQR$. This is not just a calculation; it is a study of how planes orient themselves in space.

Normal Vectors

When we talk about the angle between two planes, we are talking about the dihedral angle. Trying to measure this by looking at the edges is like trying to measure the slope of a mountain by looking at a single rock.

Instead, we use the normal vector—a vector that stands perfectly perpendicular to the plane. If we know the direction of the 'finger' pointing out of each face, the angle between those two fingers is exactly the angle between the faces themselves.

To find this 'finger' (the normal vector $\vec{n}$), we use the cross product of two vectors that lie on the plane.

The First Face $OPQ$

Let us focus on face $OPQ$. Since it passes through the origin $O(0,0,0)$, the vectors $\overrightarrow{OP}$ and $\overrightarrow{OQ}$ are simply the coordinates of $P$ and $Q$.

So, $\overrightarrow{OP}=\hat{i}+2\hat{j}+\hat{k}$ and $\overrightarrow{OQ}=2\hat{i}+\hat{j}+3\hat{k}$. To find the normal $\overrightarrow{n_1}$, we compute the cross product:

Expanding this determinant, we get:

This vector $\overrightarrow{n_1}$ is the orientation of our first face.

The Second Face $PQR$

Now, for face $PQR$, we need two vectors on the plane. Let us use $\overrightarrow{PQ}$ and $\overrightarrow{PR}$.

Calculating these: $\overrightarrow{PQ}=(2-1)\hat{i}+(1-2)\hat{j}+(3-1)\hat{k}=\hat{i}-\hat{j}+2\hat{k}$ and $\overrightarrow{PR}=(-1-1)\hat{i}+(1-2)\hat{j}+(2-1)\hat{k}=-2\hat{i}-\hat{j}+\hat{k}$.

Now, we find the normal $\overrightarrow{n_2}$ by taking the cross product:

Expanding this, we get:

The Final Convergence

We now have our two normal vectors: $\overrightarrow{n_1}=5\hat{i}-\hat{j}-3\hat{k}$ and $\overrightarrow{n_2}=\hat{i}-5\hat{j}-3\hat{k}$. The angle $\theta$ between the planes is given by:

First, the dot product: $\overrightarrow{n_1}\cdot \overrightarrow{n_2}=(5)(1)+(-1)(-5)+(-3)(-3)=5+5+9=19$.

Next, the magnitudes:

Substituting these into our formula, we get:

Thus, the final angle is $\theta ={\cos }^{-1}\left(\frac{19}{35}\right)$. We have successfully navigated the 3D geometry of the tetrahedron and arrived at the solution with elegance and precision.