Given the relationships a×c=b and a⋅c=3, we need to find a⋅(c×b−b−c). First, consider: (a×c)2+(a⋅c)2=|a|2⋅|c|2 We are given |b|2=(a×c)2. Thus: |b|2+9=|a|2⋅|c|2 Start with: 27+9=6⋅|c|2⟹36=6⋅|c|2⟹|c|2=6 We then know: |c|2a−(a⋅c)c=c×b Substituting: 6a−3c=c×b Now compute: a⋅(6a−3c−b−c) Breaking it down: ‌6a2=6×|a|2=6×6 ‌−4(a⋅c)=−4×3=−12
a⋅b=0 because a×c=b implies orthogonality of a and b. Substituting: =6×6−12−0=36−12−0=24 Thus, the final result is: 24