Question

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

Answer 1

Answer: 3

PLAN If $a,b,c$ are any three vectors
Then $|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}{|}^2\ge 0$
$\Rightarrow |\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2+|\overrightarrow{c}{|}^2+2(\overrightarrow{a}\cdot \overrightarrow{b}+\overrightarrow{b}\cdot \overrightarrow{c}+\overrightarrow{c}\cdot \overrightarrow{a})\ge 0$
$\therefore \overrightarrow{a}\cdot \overrightarrow{b}+\overrightarrow{b}\cdot \overrightarrow{c}+\overrightarrow{c}\cdot \overrightarrow{a}\ge \frac{-1}{2}\left(|\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2+|\overrightarrow{c}{|}^2\right)$
Given, $|\overrightarrow{a}-\overrightarrow{b}{|}^2+|\overrightarrow{b}-\overrightarrow{c}{|}^2+|\overrightarrow{c}-\overrightarrow{a}{|}^2=9$
$\Rightarrow |\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2-2\overrightarrow{a}\cdot \overrightarrow{b}+|\overrightarrow{b}{|}^2+|\overrightarrow{c}{|}^2-2\overrightarrow{b}\cdot \overrightarrow{c}+|\overrightarrow{c}{|}^2+|\overrightarrow{a}{|}^2$
$-2\overrightarrow{c}\cdot \overrightarrow{a}=9$
$\Rightarrow 6-2(\overrightarrow{a}\cdot \overrightarrow{b}+\overrightarrow{b}\cdot \overrightarrow{c}+\overrightarrow{c}\cdot \overrightarrow{a})=9[\because |\overrightarrow{a}|=|\overrightarrow{b}|=|\overrightarrow{c}|=1]$
$\Rightarrow \overrightarrow{a}\cdot \overrightarrow{b}+\overrightarrow{b}\cdot \overrightarrow{c}+\overrightarrow{c}\cdot \overrightarrow{a}=-\frac{3}{2}.....$(i)
Also, $\overrightarrow{a}\cdot \overrightarrow{b}+\overrightarrow{b}\cdot \overrightarrow{c}+\overrightarrow{c}\cdot \overrightarrow{a}\ge \frac{-1}{2}\left(|\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2+|\overrightarrow{c}{|}^2\right)$
$\ge -\frac{3}{2}......$(ii)
From Eqs. (i) and (ii), $|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|=0$
as $\overrightarrow{a}\cdot \overrightarrow{b}+\overrightarrow{b}\cdot \overrightarrow{c}+\overrightarrow{c}\cdot \overrightarrow{a}$ is minimum when $|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|=0$
$\Rightarrow \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0$
$\therefore |2a+5b+5c|=|2a+5(b+c)|=|2a-5a|=3$