Animated Solution for Mathematics - Vector Algebra: If four distinct points with position vectors $\vec{a},\vec{b},\vec{c}$ and $\vec{d}$ are coplanar, then $[\vec{a}\vec{b}\vec{c}]$ is equal to

Welcome, future engineer. Today, we are not just solving a vector problem; we are exploring the very nature of 3D space. When you see the word 'coplanar' in a JEE Advanced problem, I want you to stop and visualize.

Imagine you are standing in a vast, empty room. You have a perfectly flat, infinite sheet of paper floating in the air. Four points, $A,B,C,$ and $D$, are marked on this sheet. They are not floating above or below; they are locked onto this surface. This is the essence of coplanarity.

To translate this geometric reality into the language of mathematics, we need a common anchor. Let us pick point $A$ as our origin. From this anchor, we can draw three vectors: $\overrightarrow{AB}$, $\overrightarrow{AC}$, and $\overrightarrow{AD}$.

Mathematically, these are represented by the position vectors $\vec{b}-\vec{a}$, $\vec{c}-\vec{a}$, and $\vec{d}-\vec{a}$. Because these points are coplanar, these three vectors must also lie on the same plane.

In the world of vectors, the test for coplanarity is the Scalar Triple Product (STP). Why? Because the STP, denoted as $[\vec{u}\vec{v}\vec{w}]$, calculates the volume of the parallelepiped formed by those vectors. If the vectors are coplanar, they cannot form a 3D shape; they have no 'height' relative to each other. Thus, the volume must be zero. We write this as:

Taming the Algebra

Now, I know what you are thinking. "This looks like a nightmare to expand." But let us take a breath. We use the definition of the box product: $[\vec{u}\vec{v}\vec{w}]=\vec{u}\cdot (\vec{v}\times \vec{w})$.

Our equation becomes:

Let us focus on the cross product first. Distributing the terms, we get $(\vec{c}\times \vec{d})-(\vec{c}\times \vec{a})-(\vec{a}\times \vec{d})+(\vec{a}\times \vec{a})$.

Here is the beauty of vector algebra: $\vec{a}\times \vec{a}$ is zero because any vector crossed with itself is the null vector. Furthermore, we can use the anti-commutative property of the cross product to clean this up: $-\vec{c}\times \vec{a}$ becomes $\vec{a}\times \vec{c}$, and $-\vec{a}\times \vec{d}$ becomes $\vec{d}\times \vec{a}$. Our expression is now much cleaner:

Now, we dot this with $(\vec{b}-\vec{a})$. Distributing the dot product gives us six terms. When we convert these back into box product notation, we get:

Look closely at the last two terms: $[\vec{a}\vec{a}\vec{c}]$ and $[\vec{a}\vec{d}\vec{a}]$. They contain repeated vectors! In any box product, if two vectors are identical, the volume is zero. They vanish into thin air. We are left with:

We want to isolate $[\vec{a}\vec{b}\vec{c}]$. Notice that $[\vec{b}\vec{a}\vec{c}]=-[\vec{a}\vec{b}\vec{c}]$. Rearranging our equation, we get:

Using the cyclic property, we can rewrite these to match our options perfectly. $[\vec{b}\vec{c}\vec{d}]$ becomes $[\vec{c}\vec{d}\vec{b}]$, and $-[\vec{a}\vec{c}\vec{d}]$ becomes $[\vec{d}\vec{c}\vec{a}]$.

The final result is:

You see? It wasn't a nightmare; it was a dance of symmetry. Keep this mindset, and no vector problem will ever intimidate you again.