Q.

Let a→=2i^–3j^+k^, b→=3i^+2j^+5k^ and a vector c→ be such that (a→–c→)×b→=–18i^–3j^+12k^ and a→·c→=3. If b→×c→=d→, then |a→·d→| is equal to :          [2025]

1 12  
2 18  
3 15  
4 9  

Ans.

(3)

a→×b→=|i^j^k^2–31325|

               =i^(–15–2)–j^(10–3)+k^(4+9)=–17i^–7j^+13k^

Consider, (a→×b→)–((a→–c→)×b→)

              =(a→×b→)–(a→×b→–c→×b→)=–d→          [∵ b→×c→=d→]

∴  d→=(a→–c→)×b→–(a→×b→)

              =–18i^–3j^+12k^–(–17i^–7j^+13k^)=–i^+4j^–k^

∴  a→·d→=–2–12–1=–15 ⇒|a→·d→|=|–15|=15.