Animated Solution for Mathematics - Vector Algebra: The scalar $\vec{A}\cdot (\vec{B}+\vec{C})\times (\vec{A}+\vec{B}+\vec{C})$ equals :

Welcome, future engineers! Today, we are going to dismantle a problem that looks like a daunting algebraic beast but is actually a beautiful, elegant dance of geometry.

When you first look at the expression $\vec{A}\cdot ((\vec{B}+\vec{C})\times (\vec{A}+\vec{B}+\vec{C}))$, it is natural to feel a bit overwhelmed. You might be tempted to start expanding everything immediately, but I want you to pause.

Take a deep breath. In the JEE Advanced, the most powerful tool you have is not your calculator or your speed, but your ability to visualize the physics and mathematics behind the symbols.

Seeing the Invisible

Let us start by looking at the Scalar Triple Product (STP) $\vec{a}\cdot (\vec{b}\times \vec{c})$. Geometrically, this represents the volume of a parallelepiped formed by the vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$.

If the volume is zero, it means the parallelepiped has collapsed into a flat 2D shape, which implies the vectors are coplanar. Now, look at our expression: $\vec{A}\cdot ((\vec{B}+\vec{C})\times (\vec{A}+\vec{B}+\vec{C}))$.

Let us define ${\vec{V}}_1=\vec{B}+\vec{C}$ and ${\vec{V}}_2=\vec{A}+\vec{B}+\vec{C}$. Notice that ${\vec{V}}_2$ is simply $\vec{A}+{\vec{V}}_1$.

This is the key! Because ${\vec{V}}_2$ is a linear combination of $\vec{A}$ and ${\vec{V}}_1$, all three vectors—$\vec{A}$, ${\vec{V}}_1$, and ${\vec{V}}_2$—must lie in the same plane.

They are coplanar. If they are coplanar, the volume of the parallelepiped they form is zero. Just like that, we have the answer without writing a single equation.

The Safety Net

Let us expand the cross product term: $(\vec{B}+\vec{C})\times (\vec{A}+\vec{B}+\vec{C})$. Using the distributive property, we expand this into six terms:

Now, let us apply the properties of the cross product. First, any vector crossed with itself is the zero vector. So, $\vec{B}\times \vec{B}=\vec{0}$ and $\vec{C}\times \vec{C}=\vec{0}$.

This simplifies our expression significantly to $(\vec{B}\times \vec{A})+(\vec{B}\times \vec{C})+(\vec{C}\times \vec{A})+(\vec{C}\times \vec{B})$.

Next, remember the anti-commutative property: $\vec{C}\times \vec{B}=-(\vec{B}\times \vec{C})$. Look at our expression again; the terms $(\vec{B}\times \vec{C})$ and $(\vec{C}\times \vec{B})$ cancel each other out perfectly.

We are left with only two terms: $(\vec{B}\times \vec{A})+(\vec{C}\times \vec{A})$.

Now, we bring back the outer dot product with $\vec{A}$. We need to calculate $\vec{A}\cdot ((\vec{B}\times \vec{A})+(\vec{C}\times \vec{A}))$.

Distributing the dot product, we get:

In the language of Scalar Triple Products, this is written as $[\vec{A}\vec{B}\vec{A}]+[\vec{A}\vec{C}\vec{A}]$.

In any STP, if two vectors are identical, the volume is zero because the parallelepiped has no height. Since $\vec{A}$ is repeated in both terms, both $[\vec{A}\vec{B}\vec{A}]$ and $[\vec{A}\vec{C}\vec{A}]$ are zero.

The final result is $0$.

You see? Whether you use the geometric intuition of coplanarity or the algebraic rigor of expansion, the answer remains the same. This is the beauty of physics and mathematics—everything is consistent. Keep practicing, keep visualizing, and most importantly, keep falling in love with the logic behind the numbers. You have got this!