Animated Solution for Mathematics - Vector Algebra: Consider the set of eight vectors $V=\{a\hat{i}+b\hat{j}+c\hat{k}:a,b,c\in \{-1,1\}\}$. Three non-coplanar vectors can be chosen from $V$ in $2^p$ ways. Then $p$ is

The set $V=\{a\hat{i}+b\hat{j}+c\hat{k}:a,b,c\in \{-1,1\}\}$ represents the eight vertices of a cube centered at the origin $(0,0,0)$. Each component $a,b,c$ has two possible values, leading to a total of:

Visualizing these points as vertices of a cube allows us to leverage the inherent symmetry of the structure. This geometric perspective is essential for identifying the constraints of the problem.

We must select three vectors from these eight such that they are non-coplanar. A critical constraint arises from the four body diagonals of the cube. For every vector $\vec{v}\in V$, its negative $-\vec{v}$ is also in $V$.

These pairs, such as $(1,1,1)$ and $(-1,-1,-1)$, are collinear as they lie on the same line passing through the origin. If we select two vectors from the same body diagonal, they are collinear, which forces any third vector to lie in the same plane as that line.

To ensure the three vectors are non-coplanar, we must strictly avoid picking two vectors from the same body diagonal.

To satisfy the non-coplanarity condition, we must select three vectors from three distinct body diagonals. We have four body diagonals available, and we must choose three of them. The number of ways to choose these diagonals is:

For each of the three chosen diagonals, we must select exactly one of the two available vectors. This provides $2$ choices for each of the three diagonals, resulting in:

To find the total number of successful combinations, we multiply the number of ways to choose the diagonals by the number of ways to choose the vectors within those diagonals:

The problem states that this value is equal to $2^p$. Therefore, we set up the equation:

Since $32=2^5$, we conclude that $p=5$. This result demonstrates how geometric insight and combinatorial logic simplify complex spatial problems.