Let →a and →b are two non-zero, non-collinear vectors (|→a|=1) such that vectors 3(→a×→b) and 2(→b−(→a.→b)→a) represents two sides of a triangle. If area of triangle is 34(|→b|4+4) then the value of |→b|2 is
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Solution
→v1=3(→a×→b) →v2=2(→b−(→a.→b)→a) →v1.→v2=0 |→v1|=3|→a||→b|sinθ |→v2|=2√(→b−(→a.→b)→a)2 =2√|→b|2−2(→a.→b)2+(→a.→b)2 =2√|b|2−(→a.→b)2 =2|→b|sinθ Area=12.3|→b|sinθ.2|→b|sinθ =34(|→b|4+4|) ⇒|→b|2sin2θ=14(|→b|4+4) 4sin2θ=|→b|2+4|→b|2≥4 using AM ≥ GM So equality holds because sin2θ=1 |→b|4=4 |→b|2=2