Animated Solution for Mathematics - Vector Algebra: If $\vec{a},\vec{b},\vec{c}$ are non coplanar vectors and $\lambda$ is a real number then $[\lambda (\vec{a}+\vec{b}){\lambda }^2\vec{b}\lambda \vec{c}]=[\vec{a}\vec{b}+\vec{c}\vec{b}]$ for

Unlocking the Scalar Triple Product

Welcome, fellow traveler in the realm of JEE Advanced mathematics. Today, we are not just solving an equation; we are exploring the architecture of three-dimensional space.

We are given a problem involving vectors $\vec{a},\vec{b},\vec{c}$ and a mysterious scalar $\lambda$. At first glance, the equation

might look like a chaotic jumble of symbols. But I want you to take a deep breath. In physics and mathematics, complexity is often just simplicity in disguise. Let us peel back the layers.

The Non-Coplanar Key

The problem begins with a crucial piece of information: $\vec{a},\vec{b},\vec{c}$ are non-coplanar. Why does this matter? Imagine these three vectors as the edges of a box—a parallelepiped.

If they were coplanar, they would lie flat on a single sheet of paper, and the volume of that box would be zero. But because they are non-coplanar, they reach out into the third dimension, creating a solid volume.

In the language of vectors, this means their scalar triple product, denoted as $[\vec{a}\vec{b}\vec{c}]$, is strictly non-zero. This is our 'get out of jail free' card. It allows us to perform algebraic operations that would otherwise be forbidden, like dividing by the product itself.

The LHS—The Power of Linearity

Let us look at the Left Hand Side (LHS): $[\lambda (\vec{a}+\vec{b}){\lambda }^2\vec{b}\lambda \vec{c}]$. We have scalars $\lambda ,{\lambda }^2,\lambda$ multiplying our vectors.

One of the most elegant properties of the scalar triple product is its linearity. We can extract these scalars out of the box just like we factor out constants from an integral or a derivative.

So, we pull out $\lambda$ from the first position, ${\lambda }^2$ from the second, and $\lambda$ from the third. Multiplying these together, we get $\lambda \cdot {\lambda }^2\cdot \lambda ={\lambda }^4$. Now, we are left with ${\lambda }^4[\vec{a}+\vec{b}\vec{b}\vec{c}]$.

But wait, we can go further! The scalar triple product is distributive. We can split the sum $\vec{a}+\vec{b}$ into two separate boxes:

Look closely at that second term, $[\vec{b}\vec{b}\vec{c}]$. It contains two identical vectors. Geometrically, if you try to build a box where two edges are the same, the box collapses into a flat plane. The volume is zero! Thus, the second term vanishes, and our LHS simplifies beautifully to ${\lambda }^4[\vec{a}\vec{b}\vec{c}]$.

The RHS—The Trap of Order

Now, let us turn our attention to the Right Hand Side (RHS): $[\vec{a}\vec{b}+\vec{c}\vec{b}]$. We apply the same distributive logic.

We split the middle term to get $[\vec{a}\vec{b}\vec{b}]+[\vec{a}\vec{c}\vec{b}]$. Again, the first term $[\vec{a}\vec{b}\vec{b}]$ contains repeated vectors, so it becomes zero. We are left with $[\vec{a}\vec{c}\vec{b}]$.

Here is where many students stumble. We want to compare this to our LHS, which is in the order $[\vec{a}\vec{b}\vec{c}]$. Our RHS is $[\vec{a}\vec{c}\vec{b}]$.

They are almost the same, but the order of $\vec{b}$ and $\vec{c}$ is swapped. Remember the fundamental rule of the scalar triple product: swapping any two adjacent vectors flips the sign of the product. So, $[\vec{a}\vec{c}\vec{b}]=-[\vec{a}\vec{b}\vec{c}]$. The RHS is simply $-[\vec{a}\vec{b}\vec{c}]$.

The Final Confrontation

We have arrived at the climax of our journey. We equate our simplified LHS and RHS:

Because we established in Phase 1 that the vectors are non-coplanar, we know for a fact that $[ \vec{a} \ \vec{b} \ \vec{c} ] eq 0$. This allows us to divide both sides by the scalar triple product without any fear. We are left with the stark, simple equation: ${\lambda }^4=-1$.

Now, think about the nature of real numbers. If you take any real number $\lambda$ and raise it to the fourth power, you are essentially squaring its square. Since the square of any real number is non-negative, the fourth power must also be non-negative.

It is impossible for ${\lambda }^4$ to be $-1$ if $\lambda$ is a real number. Therefore, there are no real values of $\lambda$ that satisfy this equation. We have navigated the complexity, simplified the vectors, and arrived at a definitive, logical conclusion.