Animated Solution for Mathematics - Vector Algebra: If $\vec{a}\cdot \vec{b}=1,\vec{b}\cdot \vec{c}=2$ and $\vec{c}\cdot \vec{a}=3$, then the value of $[\vec{a}\times (\vec{b}\times \vec{c}),\vec{b}\times (\vec{c}\times \vec{a}),\vec{c}\times (\vec{b}\times \vec{a})]$ is:

Welcome, future engineers! Today, we are going to tackle a problem that looks like a monster but is actually a masterclass in elegance. We are asked to evaluate the scalar triple product of three complex vector expressions: $[\vec{a}\times (\vec{b}\times \vec{c}),\vec{b}\times (\vec{c}\times \vec{a}),\vec{c}\times (\vec{b}\times \vec{a})]$.

When you first see this, it is natural to feel a bit intimidated. It looks like a chaotic mess of cross products. But in physics and mathematics, complexity is often just a mask for underlying simplicity. Let's peel back that mask together.

The Vector Triple Product

Before we dive into the algebra, we need our most powerful tool: the Vector Triple Product (VTP) identity. This is our 'decoder ring' for nested cross products.

The identity states that for any three vectors $\vec{x},\vec{y},\vec{z}$:

We call this the BAC-CAB rule. It is a lifesaver because it transforms a cross product—which is geometrically complex—into a simple linear combination of vectors. Our mission is to apply this rule to each of the three terms in our target expression.

Let's define our three vectors as $\vec{u},\vec{v},$ and $\vec{w}$. First, consider $\vec{u}=\vec{a}\times (\vec{b}\times \vec{c})$. Applying the BAC-CAB rule, we get:

Given that $\vec{a}\cdot \vec{c}=3$ and $\vec{a}\cdot \vec{b}=1$, this simplifies beautifully to:

Next, let's tackle $\vec{v}=\vec{b}\times (\vec{c}\times \vec{a})$. Again, applying the VTP identity:

Substituting $\vec{b}\cdot \vec{a}=1$ and $\vec{b}\cdot \vec{c}=2$, we get:

Finally, for $\vec{w}=\vec{c}\times (\vec{b}\times \vec{a})$:

Substituting $\vec{c}\cdot \vec{a}=3$ and $\vec{c}\cdot \vec{b}=2$, we arrive at:

Now, look at what we have created. We have transformed three terrifying cross-product expressions into three simple linear combinations:

1. $\vec{u}=3\vec{b}-\vec{c}$ 2. $\vec{v}=\vec{c}-2\vec{a}$ 3. $\vec{w}=3\vec{b}-2\vec{a}$

Do you see the hidden harmony? If we add $\vec{u}$ and $\vec{v}$ together, look at what happens to the $\vec{c}$ terms:

This is exactly $\vec{w}$! We have discovered that $\vec{u}+\vec{v}=\vec{w}$.

The Geometry of Zero

In vector algebra, if one vector can be expressed as a linear combination of others, those vectors are linearly dependent. Geometrically, this means that $\vec{u},\vec{v},$ and $\vec{w}$ all lie in the same plane—they are coplanar.

The scalar triple product $[\vec{u},\vec{v},\vec{w}]$ represents the volume of the parallelepiped formed by these three vectors. If the vectors are coplanar, the parallelepiped has no height; it is essentially a flat, 2D shape.

Therefore, its volume must be zero. And there you have it! Through the power of the VTP identity and a keen eye for linear relationships, we have dismantled a complex problem and found the elegant truth hidden within.