Animated Solution for Mathematics - Vector Algebra: If $\vec{a},\vec{b}$ and $\vec{c}$ are three non coplanar vectors, then $(\vec{a}+\vec{b}+\vec{c})\cdot [(\vec{a}+\vec{b})\times (\vec{a}+\vec{c})]$ equals

Imagine you are standing in a three-dimensional room, holding three vectors $\vec{a},\vec{b},$ and $\vec{c}$. Because they are non-coplanar, they do not lie flat on the floor; they reach out into the space around you, defining the edges of a parallelepiped.

This is the physical soul of the Scalar Triple Product. When we see an expression like $(\vec{a}+\vec{b}+\vec{c})\cdot [(\vec{a}+\vec{b})\times (\vec{a}+\vec{c})]$, it is easy to feel overwhelmed by the algebra. But let us take a breath and break this down, step by step.

We begin by looking at the inner term: $(\vec{a}+\vec{b})\times (\vec{a}+\vec{c})$. Just like in basic algebra, the cross product is distributive.

We multiply each term in the first bracket with each term in the second, but we must be careful—the cross product is anti-commutative, meaning $\vec{u}\times \vec{v}=-(\vec{v}\times \vec{u})$. We expand it as:

Here is where the beauty of geometry simplifies our work. Look at the first term: $\vec{a}\times \vec{a}$.

The angle between a vector and itself is $0^{\circ }$. Since the cross product magnitude involves $\sin (\theta )$, and $\sin (0)=0$, this term vanishes into the zero vector $\vec{0}$.

Our expression is now:

The complexity is already melting away.

Now, we bring back the outer term: $(\vec{a}+\vec{b}+\vec{c})\cdot [(\vec{a}\times \vec{c})+(\vec{b}\times \vec{a})+(\vec{b}\times \vec{c})]$. We are essentially calculating the volume of a new parallelepiped formed by these combinations.

When we distribute the dot product, we generate terms that look like $\vec{u}\cdot (\vec{v}\times \vec{w})$, which is the definition of the Scalar Triple Product $[\vec{u}\vec{v}\vec{w}]$.

A crucial property of the Scalar Triple Product is that if any two vectors are identical, the volume is zero. For example:

By applying this, we eliminate all terms where vectors repeat. We are left with only three non-zero terms:

Finally, we standardize these terms to match the cyclic order $[\vec{a}\vec{b}\vec{c}]$. Swapping any two adjacent vectors flips the sign.

Thus, $[\vec{b}\vec{a}\vec{c}]=-[\vec{a}\vec{b}\vec{c}]$ and $[\vec{c}\vec{b}\vec{a}]=-[\vec{a}\vec{b}\vec{c}]$. Substituting these back, we get:

The first two terms cancel out, leaving us with the elegant final result:

We started with a daunting expression and arrived at a simple, profound truth. This is the power of vector algebra.