Let `hata,vecb and vecc` be the non-coplanar unit vectors. The angle between `hatb and hatc is alpha "between" hatc and hata is beta and "between" hata and hatb is gamma`. If `A(hatacos alpha,0),B(hatbcosbeta,0) and C(hatc cosgamma,0),` then show that in triangle ABC, `(|hataxx(hatbxxhatca)|)/(sinA)=(|hatbxx(hatcxxhata)|)/sinB = (|hatcxx(hataxxhatb)|)/sinC`

Explanation

From the sine rule, we get
`(AB)/(sin C)=(AC)/(sinB)=(BC)/(sinA)= ((AB)(BC)(CA))/(2DeltaABC)`
`BC=|vec(BC)|=|hatc cos gamma=-hatbcosbeta|=|(hata.hatb)hatc-(hatc.hata)hatb|=|(hataxx(hatbxxhatc))|`
`AC = |vec(AC)|=|hatbxx(hatcxxhata)|and AB = |vec(AB)|=hatcxx(hataxx hatb)|`
`DeltaABC=1/2|vec(BC)xxvec(BA)|`
`=1/2 |(hatc cosgamma-hatb cos beta)xx(hata cosalpha-hatbcosbeta)|`
`=1/2 |(hatc xxhata)cosalpha cosgamma+(hatbxxhatc)cosalphacosbeta+(hata xx hatb)cos beta cos alpha|`
`2DeltaABC=|sumhatn_(1)sinalphacosbeta cosgamma|`
`(|hataxx(hatbxxhatc)|)/sinA=(|hatbxx(hatcxxhata)|)/sinB=(|hatcxx(hataxxhatb)|)/sin C = (prod|hata xx(hatbxx hatc)|)/(|sum sinalpha cosbeta cosgamma hatn_(1)|)`