Let a→=i^+2j^+k^,b→=3(i^−j^+k^) Let c→ be the vector such that a→×c→=b→ and a→.c→=3.
Then a→.((c→×b→)−b→−c→) is equal to:
36
24
20
32
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(a→×c→)b→=b→.b→
[a→c→b→]=9(1+1+1)=27
a→.b→=3(1−2+1)=0⇒a→.c→=3
a→.((c→×b→)−b→−c→)=[a→c→b→]−a→.b→−a→.c→
= 27 – 0 – 3 = 24