Animated Solution for Mathematics - Vector Algebra: Which of the following expressions are meaningful?

Welcome, future engineers! Today, we are not just solving a problem; we are learning the grammar of the universe. In physics and engineering, vectors are our primary tools for describing the world—forces, velocities, and fields.

But just as you cannot write a sentence without following the rules of grammar, you cannot perform vector algebra without respecting the 'type' of your operations. This problem is a classic JEE trap designed to test whether you truly understand the soul of vector operations or if you are just blindly manipulating symbols.

Scalars and Vectors

Before we dive into the expressions, let us define our players. We have two main operations: the dot product ($\cdot$) and the cross product ($\times$).

The dot product is the 'scalar-maker.' It takes two vectors and collapses them into a single scalar value, representing how much they align.

The cross product is the 'vector-maker.' It takes two vectors and produces a new vector, perpendicular to the plane they span. Finally, we have scalar multiplication, where a simple number scales a vector's magnitude. These are the only rules that matter.

Let us look at our first candidate: $\vec{u}\cdot (\vec{v}\times \vec{w})$. We start inside the parentheses. The cross product $(\vec{v}\times \vec{w})$ gives us a vector.

Now, we have $\vec{u}\cdot (\text{vector})$. Since the dot product is defined between two vectors, this is perfectly valid!

This is the famous Scalar Triple Product, which geometrically represents the volume of a parallelepiped. It is a beautiful, meaningful quantity.

Now, consider the second candidate: $(\vec{u}\cdot \vec{v})\cdot \vec{w}$. Inside the parentheses, $(\vec{u}\cdot \vec{v})$ yields a scalar.

Now, we are left with $(\text{scalar})\cdot \vec{w}$. Here is where the trap snaps shut.

The dot product operator is a strict gatekeeper; it refuses to accept a scalar as an input. It demands a vector. Therefore, this expression is mathematically meaningless.

Next, we have $(\vec{u}\cdot \vec{v})\vec{w}$. This looks similar to the previous one, but look closely—there is no dot symbol!

The scalar $(\vec{u}\cdot \vec{v})$ is simply sitting next to the vector $\vec{w}$. This is scalar multiplication.

We are just scaling the vector $\vec{w}$ by a number. This is perfectly valid and results in a new vector.

Finally, let us examine $\vec{u}\times (\vec{v}\cdot \vec{w})$. Inside the parentheses, $(\vec{v}\cdot \vec{w})$ gives us a scalar.

We are left with $\vec{u}\times (\text{scalar})$. Just like the dot product, the cross product is a machine that only accepts two vectors.

It cannot operate on a scalar. Thus, this expression is also mathematically meaningless.

Mastery of the Fundamentals

By carefully checking the 'type' of each operation, we have successfully identified that only the first and third expressions are meaningful. The second and fourth are traps that violate the fundamental definitions of vector algebra.

Remember, in the JEE, it is not just about calculating the answer; it is about understanding the logic behind the symbols. Keep this clarity, and you will never fall for these traps again!