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Question

If a and b are unit vectors such that |a+b|=1, then find the value of |ab|

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Solution

Consider the problem

|a|=b=1a+b=1a+b2=1(a+b).(a+b)=1(a.a=|a|2)a.a+a.b+b.a+b.b=1|a|2+2a.b+b2=1(a.b=b.a)1+2a.b+1=12a.b=1a.b=12...(1)

Now

ab2=(ab).(ab)=a.aa.bb.a+b.b=|a|22a.b+b2=12×12+1(from(1))=1+1+1=3

ab=3

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