Let ` hat a , hat b` ,and `hat c` be the non-coplanar unit vectors. The angle between ` hat b` and `hat c` is `alpha` , between ` hat c` and ` hat a` is `beta` and between ` hat a` and `hat b` is `gamma` . If `A( hat a cosalpha, 0),B( hat bcosbeta, 0)` and `C( hat c cosgamma, 0),` then show that in triangle `AB C ,` `(|hat axx(hat bxx hat c)|)/(sinA)=(|hat bxx(hat cxx hat a)|)/(sinB)=(|hat cxx(hat axx hat b)|)/(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)|)`