Let a→=4i^–j^+k^, b→=11i^–j^+k^ and c→ be a vector that (a→+b→)×c→=c→×(–2a→+3b→). If (2a→+3b→)·c→=1670, then |c→|2 is equal to : [2024]
(2)
We have, a→=4i^–j^+k^ and b→=11i^–j^+k^
a→·b→=44+1+1=46, |a→|2=18, |b→|2=123
(a→+b→)×c→=c→×(–2a→+3b→) ... (i) [Given]
and (2a→+3b→)·c→=1670 ... (ii)
∴ (a→+b→)×c→=(2a→–3b→)×c→ [From (i)]
⇒ (–a→+4b→)×c→=0 ⇒ c→=λ(4b→–a→)
⇒ (2a→+3b→)·λ(4b→–a→)=1670 [From (ii)]
⇒ λ(5a→·b→–2|a→|2+12|b→|2)=1670
⇒ λ=16705×46–2×18+12×123=1
Now, c→=4b→–a→=4(11i^–j^+k^)–(4i^–j^+k^)
=40i^–3j^+3k^
∴ |c→|2=1600+9+9=1618.