Let a→,b→,c→ be three vectors such that a→×b→=2(a→×c→). If |a→|=1, |b→|=4, |c→|=2, and the angle between b→ and c→ is 60°, then |a→·c→| is equal to [2026]
(3)
a→×b→-2(a→×c→)=0
a→×(b→-2c→)=0⇒b→-2c→=λa→ ...(1)
|λa→|2=|b→-2c→|2⇒λ2|a→|2=b2+4c2-4b→·c→
λ2=16+16-4·4·2·12
λ2=16
λ=±4
∵ b→-2c→=±4a→
Dot with c→⇒b→·c→-2|c→|2=±4(a→·c→)
4·2·12-2·4=±4(a→·c→)
|a→·c→|=1