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Question

Let xyz be a triangle and a=YZ,b=ZX,c=XY. If |a|=6,|c|=14,ab=30, then the values of |a×b+b×c|2100 is

A
108.00
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B
108.0
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C
108
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Solution

Since a+b+c=XY+YZ+ZX
a+b+c=0
|a+b|=|c|
36+|b|2+60=196
|b|=10
and ab=|a||b|cosθ
30=60cosθ
θ=60
Now, |a×b+b×c|2
=|a×bb×(a+b)|2
=4|a×b|2
=4|a|2|b|2sin2θ
=4×36×100×34
=10800
Hence the value of |a×b+b×c|2100 is 108.

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