Let a→,b→ and c→ be the three vectors having magnitudes 1,5 and 3, respectively, such that the angle between a→ and b→ is θ and a→×(a→×b→)=c→ Then tan θ is equal to
0
⅔
3/5
¾
a→×(a→×b→)=c→⇒|a→||a→×b→|=|c→|(∵a→⊥(a→×b→))
1(1×5)sinθ=3⇒sinθ=35⇒tanθ=34