Question

If $\hat{a},\hat{b}$ and $\hat{c}$ are unit vectors, and the maximum value of ${|2\hat{a}-3\hat{b}|}^2+{|2\hat{b}-3\hat{c}|}^2+{|2\hat{c}-3\hat{a}|}^2$ is $p,$ then the value of $\left[\frac{p}{10}\right]$ is
(Here, $[.]$ denotes the greatest integer function.)

Answer 1

Given $|\hat{a}|=|\hat{b}|=|\hat{c}|=1$
Let $y={|2\hat{a}-3\hat{b}|}^2+{|2\hat{b}-3\hat{c}|}^2+{|2\hat{c}-3\hat{a}|}^2$
$\Rightarrow y=4|\hat{a}{|}^2+9|\hat{b}{|}^2-12\hat{a}\cdot \hat{b}+4|\hat{b}{|}^2+9|\hat{c}{|}^2-12\hat{b}\cdot \hat{c}+4|\hat{c}{|}^2+9|\hat{a}{|}^2-12\hat{a}\cdot \hat{c}$
$\Rightarrow y=3(4+9)-12(\hat{a}\cdot \hat{b}+\hat{b}\cdot \hat{c}+\hat{c}\cdot \hat{a})$
$\Rightarrow y=39-12(\hat{a}\cdot \hat{b}+\hat{b}\cdot \hat{c}+\hat{c}\cdot \hat{a})$

As $|\hat{a}+\hat{b}+\hat{c}{|}^2\ge 0$
$\Rightarrow 3+2(\hat{a}\cdot \hat{b}+\hat{b}\cdot \hat{c}+\hat{c}\cdot \hat{a})\ge 0\Rightarrow \hat{a}\cdot \hat{b}+\hat{b}\cdot \hat{c}+\hat{c}\cdot \hat{a}\ge \frac{-3}{2}$
$\Rightarrow y_{max}=39+\left(12\right)\left(\frac{3}{2}\right)$
$=39+18=57$
$\therefore \left[\frac{p}{10}\right]=\left[\frac{57}{10}\right]=5$