Consider the set of eight vectors\ $V=\{a \hat{i}+b \hat{j}+c \hat{k}: a, b, c \in\{-1,1\}\}$. Thre

Question

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 $\_\_\_\_$ .\

Professor PC Thomas Classes · Answer

(5) : Let vectors be given as\ $\vec{a}=(1,1,1) \quad \vec{a}^{\prime}=(-1,-1,-1)$\ $\vec{b}=(-1,1,1) \quad \vec{b}^{\prime}=(1,-1,-1)$\ $\vec{c}=(1,-1,1) \quad \vec{c}^{\prime}=(-1,1,-1)$\ $\vec{d}=(1,1,-1) \quad \vec{d}^{\prime}=(-1,-1,1)$\ $\vec{a}$ and $\vec{a}^{\prime}$ with any of the remaining six will produce a set of three coplanar vectors.\ Number of such choices $=4 	imes 6=24$\ The number of non coplanar vectors\ $={ }^{8} C_{3}-24=56-24=32=2^{5}$\