If a→.c→=3, a→=i→+j→+k→,b→=j→-k→,a→×c→=b,→ Find a→b→c→.
Explanation of the above question:
Find a→b→c→:
Given,
a→.c→=3
a→=i→+j→+k→
a→×c→=b→
a→b→c→ =-[a→c→b→]
=-(a→×c→.b→) By formulas of the scalar triple product
=-(b→.b→) ∵a→×c→=b→
=-(j→-k→.j→-k→)
=-(2)
Hence, the value of a→b→c→ is -2.