Let a→=2i^+3j^+4k^, b→=i^-2j^-2k^ and c→=-i^+4j^+3k^. If d→ is a vector perpendicular to both b→ and c→, and a→·d→=18, then |a→×d→|2 is equal to [2023]
(1)
∵ d→ is perpendicular to both b→ and c→.
∴ d→ is parallel to plane (b→×c→) ⇒ d→=λ(b→×c→)
Now, b→×c→=|i^j^k^1-2-2-143|=i^(-6+8)-j^(3-2)+k^(4-2)
=2i^-j^+2k^ ⇒ d→=λ(2i^-j^+2k^)
⇒ a→·d→=(2i^+3j^+4k^)·(2λi^-λj^+2λk^)
⇒ 18=4λ-3λ+8λ ⇒ λ=2
Thus, d→=2(2i^-j^+2k^)=4i^-2j^+4k^
Now, a→×d→=|i^j^k^2344-24|=i^(12+8)-j^(8-16)+k^(-4-12)
=20i^+8j^-16k^⇒ |a→×d→|2=202+82+(-16)2=720