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Question

(a.ˆi)(a׈i)+(a.ˆj)(a׈j)+(.ˆk)(a׈k) is equal to

A
3a
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B
a
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C
0
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D
None of these
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Solution

The correct option is C 0
Let a=a1ˆi+a2ˆj+a3ˆk
Then, a.ˆi=a1,a.ˆj=a2,a.ˆk=a3
a׈i=a2ˆk+a3ˆj.a׈j=a1ˆka3ˆi
and a׈k=a1ˆj+a2ˆi
Therefore, (a.ˆi)(a׈i)+(a.ˆj)(a׈j)+(a.ˆk)(a׈k)
=a1(a2ˆk+a3ˆj)+a2(a1ˆka3ˆi)+a3(a1ˆj+a2ˆi)
=0

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