IB Mathematics AA HL

Consider the set of eight vectors V={aî+bĵ+ck̂:a,b,c∈{-1,1}}. Three non-coplanar vectors can be chosen from V in 2ᵖ ways. Then p is

$\text{Given: } V=\left\{a\hat{i}+b\hat{j}+c\hat{k} : a,b,c\in\{-1,1\}\right\}$

$\text{So each vector is } (\pm 1)\,\hat{i}+(\pm 1)\,\hat{j}+(\pm 1)\,\hat{k},\ \text{giving } 8 \text{ vectors}$

$\text{To find: number of ways to choose three non-coplanar vectors } = 2^{p}$

$\text{Find } p$

$\text{The } 8 \text{ vectors of } V \text{ are position vectors of a cube's vertices}$

$|\vec{v}|=\sqrt{1^2+1^2+1^2}=\sqrt{3}\quad\text{for every } \vec{v}\in V$

$\text{Antipodal pairs: } \vec{v} \text{ and } -\vec{v}$

$\pm(1,1,1),\ \pm(1,1,-1),\ \pm(1,-1,1),\ \pm(-1,1,1)$

$\text{Plan: } 1.\ \text{total ways to choose } 3 \text{ from } 8 = \binom{8}{3}$

$2.\ \text{subtract the coplanar choices}$

$3.\ \text{answer} = \text{total} - \text{coplanar}$

$\binom{8}{3}=\dfrac{8\cdot 7\cdot 6}{3\cdot 2\cdot 1}=56$

$[\vec{a}\ \vec{b}\ \vec{c}]=\vec{a}\cdot(\vec{b}\times\vec{c})=0 \iff \text{coplanar}$

$\vec{a},\ -\vec{a},\ \vec{c}\ \text{are always coplanar}\ \Rightarrow\ [\vec{a},-\vec{a},\vec{c}]=0$

$4\times 6 = 24\ \text{coplanar triples}$

$\text{no double count: two pairs would need } 4 \text{ vectors}$

$\text{Plane}(a,b)\cap V=\{\pm a,\pm b\}\ \Rightarrow\ \text{at most } 4 \text{ vectors in a plane}$

$56 - 24 = 32$

$8\times 6\times 4 = 192$

$\dfrac{192}{3!}=32$

$32 = 2^{p}\ \Rightarrow\ 2^{5}=2^{p}\ \Rightarrow\ p=5$

$\boxed{p = 5}$