Math, asked by namku, 1 year ago

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.

Attachments:

rational: the answer cannot be 6 or 12. notice that both |a-b|^2 and |b-c|^2 need not achieve their maximum values at the same time.

namku: ans given is 6

rational: answer is wrong, consider the unit vectors : <-1/sqrt(2), 1/sqrt(2)>, <-1, 0> and <1, 0>

rational: they give you |a-b|^2 + |b-c|^2 | + |c-a|^2 = 9

namku: oh ya sorry ans is 8

rational: answer has to be 9

namku: ya ya typing error sorry it is 9.

Answers

Answered by rational

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

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 : $(a+b+c)^2\ge\,0$

namku: how to continue further?

rational: i think we need to find the minimum value of that sum of dot products

rational: il update shortly.. still working

namku: okayyyy no probs

namku: sir can u solve my next sum i posted ?

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