Question

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

Answer 1

Total number of distinct vectors $={}^8{C}_3=56$

Consider the pairs
$(\hat{i}+\hat{j}+\hat{k},-\hat{i}-\hat{j}-\hat{k})$,
$(-\hat{i}+\hat{j}+\hat{k},\hat{i}-\hat{j}-\hat{k})$,
$(\hat{i}-\hat{j}+\hat{k},-\hat{i}+\hat{j}-\hat{k})$,
$(\hat{i}+\hat{j}-\hat{k},-\hat{i}-\hat{j}+\hat{k})$

Elements in each pairs are multiple of each other.
If we take any one of these pairs and any one vector from the remaining three pairs (six vectors), then these three vectors will be coplanar.
So, number of coplanar vectors $={}^4{C}_1\times {}^6{C}_1=24$

Therefore, total number of non-coplanar vectors $=56-24=32=2^5$