JEE Advance - Mathematics (2013 - Paper 1 Offline - No. 6)
Explanation
Given, the set of eight vectors
$$\mathrm{V}=\{a \hat{i}+b \hat{j}+c \hat{k}: a, b, c \in\{-1,1\}\} .$$
Now, the eight vectors are $$\hat{i}+\hat{j}+\hat{k}, \hat{i}+\hat{j}-\hat{k}$$,
$$\begin{aligned} & \hat{i}-\hat{j}+\hat{k},-\hat{i}+\hat{j}+\hat{k}, \hat{i}-\hat{j}-\hat{k}, \\ & -\hat{i}+\hat{j}-\hat{k},-\hat{i}-\hat{j}+\hat{k} \text { and }-\hat{i}-\hat{j}-\hat{k} . \end{aligned}$$
Here, $$\hat{i}+\hat{j}+\hat{k}$$ and $$-\hat{i}-\hat{j}-\hat{k}, \hat{i}+\hat{j}-\hat{k}$$ and $$-\hat{i}-\hat{j}+\hat{k}, \hat{i}-\hat{j}+\hat{k}$$ and $$-\hat{i}+\hat{j}-\hat{k},-\hat{i}+\hat{j}+\hat{k}$$ and $$\hat{i}-\hat{j}-\hat{k}$$ are collinear vectors.
$$\begin{aligned} \text{Let} \quad & \mathrm{S}_1=\{\hat{i}+\hat{j}+\hat{k},-\hat{i}-\hat{j}-\hat{k}\}, \\ & \mathrm{S}_2=\{\hat{i}+\hat{j}-\hat{k},-\hat{i}-\hat{j}+\hat{k}\}, \\ & \mathrm{S}_3=\{\hat{i}-\hat{j}+\hat{k},-\hat{i}+\hat{j}-\hat{k}\} \text { and } \\ & \mathrm{S}_4=\{-\hat{i}+\hat{j}+\hat{k}, \hat{i}-\hat{j}-\hat{k}\} \end{aligned}$$
For the set of three non - coplanar vector, we have to select three set out of $$\mathrm{S}_1, \mathrm{~S}_2, \mathrm{~S}_3, \mathrm{~S}_4$$ and select one vector in every selected set of $$\mathrm{S_1, S_2}, \mathrm{S}_3, \mathrm{~S}_4$$
$$\begin{aligned} & \Rightarrow{ }^4 C_3 \cdot{ }^2 C_1 \cdot{ }^2 C_1{ }^2 C_1=2^p \\ & \Rightarrow \quad 4.2 .2 .2=2^p \\ & \Rightarrow \quad 2^5=2^p \\ & \Rightarrow \quad p=5 \\ \end{aligned}$$
Hints :
Recall that $$\hat{i}+\hat{j}+\hat{k}$$ and $$-\hat{i}-\hat{j}-\hat{k}, \hat{i}+\hat{j}-\hat{k}$$ and $$-\hat{i}-\hat{j}+\hat{k}, \hat{i}-\hat{j}+\hat{k}$$ and $$-\hat{i}+\hat{j}-\hat{k},-\hat{i}+\hat{j}+\hat{k}$$ and $$\hat{i}-\hat{j}-\hat{k}$$ are collinear vectors.
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