Q.

Let a→=2i^+5j^–k^, b→=2i^–2j^+2k^ and c→ be three vectors such that (c→+i^)×(a→+b→+i^)=a→×(c→+i^). If a→·c→=–29, then c→·(–2i^+j^+k^) is equal to :          [2024]

1 15  
2 5  
3 12  
4 10  

Ans.

(2)

Consider a→+b→+i^

=2i^+5j^–k^+2i^–2j^+2k^+i^

=5i^+3j^+k^

Now, (c→+i^)×(a→+b→+i^)=a→×(c→+i^)

⇒ (c→+i^)×(5i^+3j^+k^)=(2i^+5j^–k^)×(c→+i^)

⇒ (c→+i^)×(5i^+3j^+k^+2i^+5j^–k^)=0

⇒ (c→+i^)×(7i^+8j^)=0

⇒ c→+i^=λ(7i^+8j^)

a→·c→+a→·i^=λa→·(7i^+8j^)

⇒ –29+2=λ(14+40) ⇒ λ=–12

∴   c→+i^=–72i^–4j^

⇒ c→=–92i^–4j^

∴  c→·(–2i^+j^+k^)=9–4=5.