Q.

Let a→ and b→ be two vectors such that |a→|=1, |b→|=4 and a→·b→=2. If c→=(2a→×b→)–3b→ and the angle between b→ and c→ is α, then 192 sin2α is equal to __________.          [2024]


Ans.

(48)

Given |a→|=1, |b→|=4 and a→·b→=2

⇒  |a→×b→|2=|a→|2|b→|2–(a→·b→)2

     |a→×b→|2=16–4=12

Now c→=2a→×b→–3b→

⇒  |c→|2=4|a→×b→|2+9|b→|2

     |c→|2=4×12+9×16=192

     c→=(2a→×b→)–3b→          [Taking dot product with b→]

     b→·c→=0–3|b→|2=–48

Let α be the angle between b→ and c→ then

cosα=b→·c→|b→||c→|=–484×83=–32 ⇒ α=5π6

then 192sin2α=192sin2(150°)=192×sin2(180°–30°)

      =192×sin230°=192×14=48.