Animated Solution for Mathematics - Vector Algebra: If $\vec{a},\vec{b},\vec{c}$ and $\vec{d}$ are distinct vectors such that $\vec{a}\times \vec{c}=\vec{b}\times \vec{d}$ and $\vec{a}\times \vec{b}=\vec{c}\times \vec{d}$. Prove that $(\vec{a}-\vec{d})\cdot (\vec{b}-\vec{c})\ne 0$ i.e. $\vec{a}\cdot \vec{b}+\vec{d}\cdot \vec{c}\ne \vec{d}\cdot \vec{b}+\vec{a}\cdot \vec{c}$.

Unlocking Geometric Truths

Welcome, future engineers! Today, we are not just solving a problem; we are embarking on a journey through the elegant language of vector algebra.

Often, when we see cross products, our minds jump to determinants or complex components. But today, I want you to pause. Look at the structure and the symmetry.

We are given two equations: $\vec{a}\times \vec{c}=\vec{b}\times \vec{d}$ and $\vec{a}\times \vec{b}=\vec{c}\times \vec{d}$. Our mission is to prove that $(\vec{a} - \vec{d}) \cdot (\vec{b} - \vec{c}) eq 0$. Let’s peel back the layers of this mystery together.

The Power of Subtraction

When you see two equations involving the same vectors, your first instinct should be to look for a relationship between them. If we subtract the second equation from the first, we are creating a new, unified expression that holds the key to the entire problem.

Subtracting gives us:

Notice what happens here. We have successfully grouped the terms. On the left, we have $\vec{a}$ appearing in both terms. On the right, we have $\vec{d}$ appearing in both.

The Art of Factorization

Now, let’s apply the distributive property of the cross product. On the left side, we can factor out $\vec{a}$:

On the right side, we have $(\vec{b}\times \vec{d})-(\vec{c}\times \vec{d})$. Factoring out $\vec{d}$ from the right requires care with the order, yielding $(\vec{b}-\vec{c})\times \vec{d}$.

Wait! We have $(\vec{c}-\vec{b})$ on one side and $(\vec{b}-\vec{c})$ on the other. Applying the anti-commutative property $\vec{u}\times \vec{v}=-(\vec{v}\times \vec{u})$, we align the terms:

Now, our equation looks like this:

The Geometric Revelation

Bring everything to one side to obtain:

Factoring out the common term $(\vec{c}-\vec{b})$, we arrive at the beautiful, concise result:

This is where the physics and geometry collide. We have the cross product of two vectors, $(\vec{a}-\vec{d})$ and $(\vec{c}-\vec{b})$, resulting in the zero vector.

Since the problem implies these vectors are non-zero, the cross product being zero means they are parallel. They are collinear in their orientation.

The Final Proof

If two vectors are parallel, their dot product cannot be zero. The dot product is defined as $\vec{u}\cdot \vec{v}=|\vec{u}||\vec{v}|\cos (\theta )$. For parallel vectors, $\theta$ is $0^{\circ }$ or $180^{\circ }$, meaning $\cos (\theta )$ is $1$ or $-1$.

Therefore:

Since $(\vec{c}-\vec{b})=-(\vec{b}-\vec{c})$, we conclude:

Expanding this, we get $\vec{a} \cdot \vec{b} - \vec{a} \cdot \vec{c} - \vec{d} \cdot \vec{b} + \vec{d} \cdot \vec{c} eq 0$. This rearranges to the final inequality:

See? It wasn't about memorizing formulas. It was about seeing the structure, respecting the properties of vectors, and following the logic to its inevitable conclusion. You have mastered this.