Animated Solution for Mathematics - Vector Algebra: Let $\vec{a},\vec{b}$ and $\vec{c}$ be three unit vectors such that $\vec{a}+\vec{b}+\vec{c}=\vec{0}$. If $\lambda =\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a}$ and $\vec{d}=\vec{a}\times \vec{b}+\vec{b}\times \vec{c}+\vec{c}\times \vec{a}$, then the ordered pair $(\lambda ,\vec{d})$ is equal to :

A Geometric Symphony

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a problem; we are peeling back the layers of vector algebra to reveal the elegant geometry hidden beneath.

When you look at the equation $\vec{a}+\vec{b}+\vec{c}=\vec{0}$, do not just see symbols. See a story. Imagine three arrows, each of unit length, chasing each other in a perfect, closed loop.

Because they are all unit vectors and they sum to zero, they are forced to form an equilateral triangle. This is the geometric soul of the problem. If you can visualize this, you have already won half the battle.

Unlocking the Dot Product

We are tasked with finding $\lambda =\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a}$. How do we extract these dot products from a simple sum? The answer lies in the power of squaring.

In vector algebra, squaring is the bridge between the sum of vectors and their dot products. We take our starting point:

When we expand this, we are essentially performing a dot product of the sum with itself. The expansion is a classic identity that you should have etched into your memory:

Look at the beauty of this structure! The term in the parentheses is exactly our $\lambda$. Since we are dealing with unit vectors, we know that $|\vec{a}|=|\vec{b}|=|\vec{c}|=1$.

Substituting these values, the equation becomes:

Solving for $\lambda$, we find $\lambda =-\frac{3}{2}$. It is clean, it is precise, and it is mathematically satisfying.

The Cross Product Transformation

Now, let us turn our attention to the vector $\vec{d}=\vec{a}\times \vec{b}+\vec{b}\times \vec{c}+\vec{c}\times \vec{a}$. This looks intimidating, but we have a secret weapon: substitution.

We know that $\vec{a}+\vec{b}+\vec{c}=\vec{0}$, which implies $\vec{c}=-(\vec{a}+\vec{b})$. By substituting this into our expression for $\vec{d}$, we reduce the complexity of the problem significantly.

Let us substitute $\vec{c}$ into the second and third terms of $\vec{d}$:

Now, we must be careful with the distributive property of the cross product. Remember, the cross product is anti-commutative, meaning $\vec{x}\times \vec{y}=-(\vec{y}\times \vec{x})$.

Expanding the middle term, $\vec{b}\times (-\vec{a}-\vec{b})$, we get:

Since $\vec{b}\times \vec{b}=\vec{0}$ and $-\vec{b}\times \vec{a}=\vec{a}\times \vec{b}$, this term simplifies beautifully to $\vec{a}\times \vec{b}$.

Similarly, expanding the last term, $(-\vec{a}-\vec{b})\times \vec{a}$, we get:

Since $\vec{a}\times \vec{a}=\vec{0}$ and $-\vec{b}\times \vec{a}=\vec{a}\times \vec{b}$, this term also simplifies to $\vec{a}\times \vec{b}$.

When we bring it all together, the expression for $\vec{d}$ becomes:

We have successfully navigated the algebra. We found $\lambda =-\frac{3}{2}$ and $\vec{d}=3(\vec{a}\times \vec{b})$.

The final result is the ordered pair:

This problem is a reminder that in physics and mathematics, complexity is often just a mask for simplicity. When you encounter a wall of variables, do not panic. Look for the symmetry, use the identities, and trust the process.