Consider the set of eight vectors V={a hat{i} | vedclass

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(A) The set $V$ consists of $8$ vectors of the form $(\pm 1, \pm 1, \pm 1)$.These vectors represent the vertices of a cube centered at the origin.Tota

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$6$

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(A) The set $V$ consists of $8$ vectors of the form $(\pm 1, \pm 1, \pm 1)$.These vectors represent the vertices of a cube centered at the origin.Total number of ways to choose $3$ vectors from $8$ is $\binom{8}{3} = \frac{8 	imes 7 	imes 6}{3 	imes 2 	imes 1} = 56$.Three vectors are coplanar if they lie on the same plane passing through the origin.For any vector $\vec{v} \in V$,its negative $-\vec{v}$ is also in $V$. If we choose a pair of opposite vectors $(\vec{v}, -\vec{v})$,any third vector $\vec{u} \in V$ will form a coplanar set with them because the plane containing $\vec{v}$ and $\vec{u}$ also contains $-\vec{v}$.There are $4$ such pairs of opposite vectors: $(\hat{i}+\hat{j}+\hat{k}, -\hat{i}-\hat{j}-\hat{k})$,$(\hat{i}+\hat{j}-\hat{k}, -\hat{i}-\hat{j}+\hat{k})$,$(\hat{i}-\hat{j}+\hat{k}, -\hat{i}+\hat{j}-\hat{k})$,and $(\hat{i}-\hat{j}-\hat{k}, -\hat{i}+\hat{j}+\hat{k})$.For each pair,there are $6$ remaining vectors. Thus,there are $4 	imes 6 = 24$ sets of $3$ coplanar vectors.Number of non-coplanar sets $= 56 - 24 = 32$.We are given that the number of ways is $2^p$,so $2^p = 32 = 2^5$.Therefore,$p = 5$.

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

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