Animated Solution for Mathematics - Vector Algebra: If $\vec{u},\vec{v},\vec{w}$ are non-coplanar vectors and $p,q$ are real numbers, then the equality $[3\vec{u}p\vec{v}p\vec{w}]-[p\vec{v}\vec{w}q\vec{u}]-[2\vec{w}q\vec{v}q\vec{u}]=0$ holds for :

Imagine you are standing in a three-dimensional room. You have three vectors, $\vec{u}$, $\vec{v}$, and $\vec{w}$, originating from the corner of the floor.

If they were coplanar, they would lie flat on the floor, and the volume they enclose would be zero. However, the problem specifies that they are non-coplanar.

This means they reach out into the third dimension, forming a parallelepiped with a definite, non-zero volume. In the language of JEE Advanced, this is our anchor: the scalar triple product $[\vec{u} \ \vec{v} \ \vec{w}] eq 0$. This is the key that will unlock the entire problem.

Before we touch the equation, let us sharpen our tools. The scalar triple product is not just a random collection of symbols; it is a highly structured operator.

We have three main properties: 1. Linearity, which allows us to pull scalar coefficients out of the box. 2. Cyclic Symmetry, which allows us to rotate the vectors without changing the value. 3. Anti-Symmetry, which forces a sign change if we swap two vectors.

Think of these as the laws of physics for our vector space.

Let us break down the equation term by term. The first term is $[3\vec{u}p\vec{v}p\vec{w}]$.

Using linearity, we extract the scalars $3$, $p$, and $p$. This gives us:

Now, consider the second term: $[p\vec{v}\vec{w}q\vec{u}]$. We pull out $p$ and $q$ to obtain $pq[\vec{v}\vec{w}\vec{u}]$.

By applying cyclic symmetry, we know this is identical to $[\vec{u}\vec{v}\vec{w}]$. So, we get:

Finally, the third term is $[2\vec{w}q\vec{v}q\vec{u}]$. We extract $2$, $q$, and $q$, leaving us with $2q^2[\vec{w}\vec{v}\vec{u}]$.

Here is the trap: the order is $\vec{w},\vec{v},\vec{u}$. This is one swap away from $\vec{u},\vec{v},\vec{w}$, so we must introduce a negative sign:

Now, we combine these into our master equation:

Factoring out the common box product, we arrive at:

Since we established that the box product is non-zero, we are left with the scalar equation:

We can factor the quadratic expression $3p^2-pq-2q^2=0$ as follows:

This implies that either $p=q$ or $3p=-2q$.

If the problem implies that $p$ and $q$ are real constants, there are infinitely many pairs $(p,q)$ satisfying this relationship. However, if the context requires a specific constraint or unique solution, one must verify the coefficients provided in the original prompt.

Based on the derived quadratic, the relationship between the variables is defined by the set of points lying on the lines $p=q$ and $3p=-2q$.