Let a→=2i^+7j^-k^, b→=3i^+5k^ and c→=i^-j^+2k^. Let d→ be a vector which is perpendicular to both a→ and b→, and c→·d→=12. Then (-i^+j^-k^)·(c→×d→) is equal to [2023]
(4)
a→=2i^+7j^-k^, b→=3i^+5k^ and c→=i^-j^+2k^
d→=λ(a→×b→)=λ|i^j^k^27-1305|=λ(35i^-13j^-21k^)
Also, c→·d→=12
⇒λ(35+13-42)=12 ⇒λ=2
∴ d→=70i^-26j^-42k^
Now, (i^+j^-k^)·(c→×d→)
=(i^+j^-k^)·|i^j^k^1-1270-26-42|
=(-i^+j^-k^)·(94i^+182j^+44k^)=-94+182-44=44