Animated Solution for Mathematics - Three Dimensional Geometry: A tetrahedron has vertices at $O(0,0,0),A(1,2,1),B(2,1,3)$ and $C(-1,1,2)$. Then the angle between the faces OAB and ABC will be

A Journey into 3D Space

Welcome, future engineer. Today, we are not just solving a problem; we are exploring the architecture of a tetrahedron.

Imagine you are standing in a 3D coordinate system. You have four points: $O(0,0,0)$, $A(1,2,1)$, $B(2,1,3)$, and $C(-1,1,2)$.

These points form a pyramid-like structure. Our mission is to find the angle between two of its faces: face $OAB$ and face $ABC$. This is a classic JEE Advanced problem that tests your ability to translate geometric intuition into vector algebra.

The Philosophy of the Normal Vector

When you are asked for the angle between two planes, your first instinct might be to look at the edges. Resist that urge!

The most elegant way to solve this is to think about the 'orientation' of the planes. Every plane has a unique direction perpendicular to it, known as the normal vector.

If you find the normal vector of face $OAB$ (let's call it ${\vec{n}}_1$) and the normal vector of face $ABC$ (let's call it ${\vec{n}}_2$), the angle between the planes is simply the angle between these two normals. It is a beautiful reduction of a 3D problem into a simple vector dot product.

Unveiling the First Face

Let us focus on face $OAB$. To find its normal, we need two vectors that lie flat on this face.

Since the origin $O$ is a vertex, the vectors $\overrightarrow{OA}$ and $\overrightarrow{OB}$ are our best friends. We calculate them as $\overrightarrow{OA}=\hat{i}+2\hat{j}+\hat{k}$ and $\overrightarrow{OB}=2\hat{i}+\hat{j}+3\hat{k}$.

Now, we invoke the cross product, the magic tool that generates a perpendicular vector. We set up the determinant:

Expanding this, we get ${\vec{n}}_1=\hat{i}(6-1)-\hat{j}(3-2)+\hat{k}(1-4)$, which simplifies to ${\vec{n}}_1=5\hat{i}-\hat{j}-3\hat{k}$. That is our first normal vector!

The Second Face

Now, we turn our attention to face $ABC$. We need two vectors on this face. Let's use point $A$ as our anchor.

We find $\overrightarrow{AB}=(2-1)\hat{i}+(1-2)\hat{j}+(3-1)\hat{k}=\hat{i}-\hat{j}+2\hat{k}$ and $\overrightarrow{AC}=(-1-1)\hat{i}+(1-2)\hat{j}+(2-1)\hat{k}=-2\hat{i}-\hat{j}+\hat{k}$.

Again, we perform the cross product to find ${\vec{n}}_2$:

Expanding this determinant gives us ${\vec{n}}_2=\hat{i}(-1+2)-\hat{j}(1+4)+\hat{k}(-1-2)$, which simplifies to ${\vec{n}}_2=\hat{i}-5\hat{j}-3\hat{k}$.

The Final Synthesis

We have our two normals: ${\vec{n}}_1=5\hat{i}-\hat{j}-3\hat{k}$ and ${\vec{n}}_2=\hat{i}-5\hat{j}-3\hat{k}$. The final step is to use the dot product formula:

First, the dot product: ${\vec{n}}_1\cdot {\vec{n}}_2=(5)(1)+(-1)(-5)+(-3)(-3)=5+5+9=19$.

Next, the magnitudes: $|{\vec{n}}_1|=\sqrt{5^2+(-1{)}^2+(-3{)}^2}=\sqrt{35}$ and $|{\vec{n}}_2|=\sqrt{1^2+(-5{)}^2+(-3{)}^2}=\sqrt{35}$.

Plugging these into our formula, we get:

Thus, the angle is $\theta ={\cos }^{-1}\left(\frac{19}{35}\right)$. You have successfully navigated the 3D geometry of the tetrahedron.