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If a, b, and c are unit vectors satisfying  $|a-b|^{2}+|b-c|^{2}+|c-a|^{2}=9$ then |2a+5b+5c| is 

Answer:

Explanation:

Concept involved if a,b,c are any three vectors

 then$|a+b+c|^{2} \geq 0$

 $\Rightarrow$    $|a|^{2}+|b|^{2}+|c|^{2}+2(a.b+b.c+c.a) \geq 0$

 $\therefore$    $a.b+b.c+c.a \geq \frac{-1}{2} (|a|^{2}+|b|^{2}+|c|^{2})$

 Sol.Given   $|a-b|^{2}+|b-c|^{2}+|c-a|^{2}=9$

 $\Rightarrow$   $|a|^{2}+|b|^{2}-2a.b+|b|^{2}+|c|^{2}-2b.c+|c|^{2}+|a|^{2}-2c.a=9$

$\Rightarrow$   $6-2(a.b+b.c+c.a)=9$    $( \because  |a|=|b|=|c|=1)$

 $\Rightarrow$    a.b+b.c+c.a=-$\frac{3}{2}$....(i)

 Also, a.b+b.c+c.a  $\geq  \frac{-1}{2} (|a|^{2}+|b|^{2}+|c|^{2}) \geq -\frac{3}{2}$....(ii)

From eq.(i) and (ii)  , we get

|a+b+c|=0

as a.b+b.c+c.a is minimum when |a+b+c|=0

 $\Rightarrow$    a+b+c=0

 $\therefore$   |2a+5b+5c|=|2a+5(b+c)|=|2a-5b|=3