Animated Solution for Mathematics - Vector Algebra: If $\vec{a},\vec{b}$ and $\vec{c}$ are unit coplanar vectors, then the scalar triple product $[2\vec{a}-\vec{b},2\vec{b}-\vec{c},2\vec{c}-\vec{a}]=$

Imagine you are standing in a vast, empty 3D space with three unit vectors: $\vec{a}$, $\vec{b}$, and $\vec{c}$. The problem states that these vectors are coplanar.

This implies that all three vectors lie perfectly flat on a single 2D plane. Consequently, the volume of the parallelepiped formed by these vectors is zero.

In the language of the Scalar Triple Product, this geometric constraint is expressed as:

We are tasked with evaluating the Scalar Triple Product of three new vectors: $[2\vec{a}-\vec{b},2\vec{b}-\vec{c},2\vec{c}-\vec{a}]$. These are linear combinations of our original coplanar set.

We utilize the property that the Scalar Triple Product of linear combinations is equal to the determinant of the coefficient matrix multiplied by the original Scalar Triple Product:

We extract the coefficients for each vector systematically: 1. For $2\vec{a}-1\vec{b}+0\vec{c}$, the coefficients are $(2,-1,0)$. 2. For $0\vec{a}+2\vec{b}-1\vec{c}$, the coefficients are $(0,2,-1)$. 3. For $-1\vec{a}+0\vec{b}+2\vec{c}$, the coefficients are $(-1,0,2)$.

This gives us the following coefficient matrix:

The expression simplifies to the product of the determinant of this matrix and the original Scalar Triple Product. Since we established that $[\vec{a},\vec{b},\vec{c}]=0$, the entire expression must vanish.

Regardless of the value of the determinant, any finite number multiplied by zero results in zero. Thus:

Geometrically, this confirms that the new vectors are also coplanar. The final answer is 0.