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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),B(hatbcosbeta) and C(hatc cosgamma), then show that in triangle ABC, (|hataxx(hatbxxhatca)|)/(sinA)=(|hatbxx(hatcxxhata)|)/sinB = (|hatcxx(hataxxhatb)|)/sinC=(prod|hataxx(hat xx hatc|))/(sumsin alpha-cosbeta. cos gammahatn_(1)) where hatn_(1)=(hatbxxhatc)/(|hatbxxhatc|),hatn_(2)=(hatcxxhata)/(|hatcxxhata|)and hatn_(3)=(hataxxhatb)/(|hataxxhatb|) |
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Answer» Solution :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)|)` |
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