Animated Solution for Mathematics - Vector Algebra: If $\vec{a}\times \vec{b}=\vec{b}\times \vec{c}=\vec{c}\times \vec{a}$ then $\vec{a}+\vec{b}+\vec{c}=$

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are uncovering a hidden harmony.

We are presented with three vectors, $\vec{a}$, $\vec{b}$, and $\vec{c}$, bound by a beautiful, cyclic condition:

At first glance, this looks like a tangled mess of cross products. However, in the world of JEE Advanced, when you see such perfect symmetry, it is rarely a coincidence. It is an invitation to look deeper.

Our goal is to find the value of $\vec{a}+\vec{b}+\vec{c}$. Dealing with three separate vectors is cumbersome.

Let us simplify our life by defining a single, powerful entity:

By giving this sum a name, we shift our focus. We are no longer chasing three separate variables; we are investigating the nature of $\vec{r}$. To find out what $\vec{r}$ represents, we will use the cross product as our primary tool.

Let us perform a surgical strike. We will take the cross product of our sum vector $\vec{r}$ with $\vec{a}$.

Consider the expression $\vec{a}\times \vec{r}$. Substituting our definition, we get:

Now, we invoke the distributive property of the cross product. We expand this to get:

Here is where the beauty reveals itself. Recall the fundamental property: the cross product of any vector with itself is the zero vector, $\vec{a}\times \vec{a}=\vec{0}$.

The first term vanishes, leaving us with:

Look at the remaining terms. We have $\vec{a}\times \vec{b}$ and $\vec{a}\times \vec{c}$. Remember that reversing the order of a cross product flips the sign:

Since the problem states $\vec{c}\times \vec{a}=\vec{a}\times \vec{b}$, it follows that $\vec{a}\times \vec{c}=-(\vec{a}\times \vec{b})$. Substituting this back into our equation:

We have discovered that $\vec{a}\times \vec{r}=\vec{0}$. This is a profound statement, as it implies that $\vec{r}$ is parallel to $\vec{a}$.

We can repeat this exact logic for $\vec{b}$. If we compute $\vec{b}\times \vec{r}$, we find that it also equals $\vec{0}$, meaning $\vec{r}$ is parallel to $\vec{b}$ as well.

Think about this: $\vec{r}$ is parallel to $\vec{a}$, and $\vec{r}$ is parallel to $\vec{b}$. Unless $\vec{a}$ and $\vec{b}$ are themselves parallel, the only vector that can be parallel to both is the zero vector.

Therefore, we conclude that:

The complexity collapses into perfect, elegant simplicity. You have navigated the logic, handled the cross products, and arrived at the truth. That is the essence of JEE mathematics—finding the order within the chaos.