Q.

Let a→=2i^–3j^+4k^, b→=3i^+4j^–5k^ and a vector c→ be such that a→×(b→+c→)+b→×c→=i^+8j^+13k^. If a→·c→=13, then (24–b→·c→) is equal to __________.          [2024]


Ans.

(46)

We have, a→×(b→+c→)+b→×c→=i^+8j^+13k^

⇒  (a→×b→)+(a→×c→)+(b→×c→)=i^+8j^+13k^

⇒  a→×(a→×b→)+a→×(a→×c→)+a→×(b→×c→)=a→×(i^+8j^+13k^)

⇒  (a→·b→)a→–a2b→+(a→·c→)a→–a2c→+(a→·c→)b→–(a→·b→)c→=a→×(i^+8j^+13k^)

⇒  –26a→–29b→+13a→–29c→+13b→+26c→=|i^j^k^2–341813|

⇒  –13a→–16b→–3c→=–71i^–22j^+19k^

⇒  –13a→·b→–16b2–3b→·c→=–213–88–95

⇒  338–800–3b→·c→=–396

⇒  –462–3b→·c→=–396 ⇒ b→·c→=–22

.∴  24–b→·c→=24+22=46