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Question

If |a|=2,|b|=3, and |c|=4, then |ab|2+|bc|2+|ca|2(12k+λ), then λ is equal to ....................

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Solution

Consider two vectors, a,b
|ab|2=a2+b22abcosθ

Now this is maximum when cosθ=1 or the vectors are anti-parallel.

In that case the above expression reduces to (a+b)2.

Hence the maximum value of
|ab|2+|bc|2+ca|2
=(a+b)2+(b+c)2+(c+a)2
=2(a2+b2+c2)+2(ab+bc+ac)
=2(4+9+16)+2(6+12+8)
=2(29)+2(26)
=2(55)
=110

Thus
|ab|2+|bc|2+|ca|2110

Now
12k+λ=110
110=108+2
=12(9)+2

Hence k=9 and λ=2.

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