I think the ans shud be 12 but the ans given is 6. cant cosangle be -1 ? then the ans will be 12. see attached q.

1
answer is wrong, consider the unit vectors : <-1/sqrt(2), 1/sqrt(2)>, <-1, 0> and <1, 0>
they give you |a-b|^2 + |b-c|^2 | + |c-a|^2 = 9
oh ya sorry ans is 8
answer has to be 9
ya ya typing error sorry it is 9.

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2015-04-18T15:31:50+05:30

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|a-b|^2+|b-c|^2+|c-a|^2
=a^2-2a\,\cdot\,b+b^2+b^2+c^2-2b\,\cdot\,c+c^2+a^2-2c\,\cdot\,a
=2(a^2+b^2+c^2-(a\cdot b+b\cdot c+ c\cdot a))

since a,b,c are unit vectors, their magnitude is 1. therefore a^2=|a|^2=1^2=1 :
=2(1+1+1-(a\,\cdot\,b+b\,\cdot\,c+ c\,\cdot\,a))
=6-2(a\,\cdot\,b+b\,\cdot\,c+ c\,\cdot\,a))
=6-((a+b+c)^2-(a^2+b^2+c^2))
=6-((a+b+c)^2-(3))
=9-(a+b+c)^2
\le9

because a square is always nonegative : (a+b+c)^2\ge\,0
2 5 2
how to continue further?
i think we need to find the minimum value of that sum of dot products
il update shortly.. still working
okayyyy no probs
sir can u solve my next sum i posted ?