Question

If $\overrightarrow{a},\overrightarrow{b}$ and $\overrightarrow{c}$ are unit vectors , then ${\left|\overrightarrow{a}-\overrightarrow{b}\right|}^2+{\left|\overrightarrow{b}-\overrightarrow{c}\right|}^2+{\left|\overrightarrow{c}-\overrightarrow{a}\right|}^2$ does not exceed

• A. $4$
• B. $9$
• C. $8$
• D. $6$

Answer 1

Answer: (B)

$9$

$|\overline{a}+\overline{b}+\overline{c}{|}^2=|\overline{a}{|}^2+|\overline{b}{|}^2+|\overline{c}{|}^2+2(\overline{a}.\overline{b}+\overline{b}.\overline{c}+\overline{a}.\overline{c})$

$\Rightarrow -2(\overline{a}.\overline{b}+\overline{b}.\overline{c}+\overline{a}.\overline{c})=|\overline{a}{|}^2+|\overline{b}{|}^2+|\overline{c}{|}^2-|\overline{a}+\overline{b}+\overline{c}{|}^2-\left(1\right)$

$|\overline{a}-\overline{b}{|}^2+|\overline{b}-\overline{c}{|}^2+|\overline{a}-\overline{c}{|}^2=|\overline{a}{|}^2+|\overline{b}{|}^2+|\overline{b}{|}^2+|\overline{c}{|}^2+|\overline{a}{|}^2+|\overline{c}{|}^2-2(\overline{a}.\overline{b}+\overline{b}.\overline{c}+\overline{a}.\overline{c})$

$=2(|\overline{a}{|}^2+|\overline{b}{|}^2+|\overline{c}{|}^2)-2(\overline{a}.\overline{b}+\overline{b}.\overline{c}+\overline{a}.\overline{c})$

$\because |\overline{a}|=|\overline{b}|=|\overline{c}|=1$

$|\overline{a}-\overline{b}{|}^2+|\overline{b}-\overline{c}{|}^2+|\overline{a}-\overline{c}{|}^2=2(1+1+1)-2(\overline{a}.\overline{b}+\overline{b}.\overline{c}+\overline{a}.\overline{c})-\left(2\right)$

For (2) to be maximum,

$2(\overline{a}.\overline{b}+\overline{b}.\overline{c}+\overline{a}.\overline{c})$ has to be minimum

From (1) $\to 2(\overline{a}.\overline{b}+\overline{b}.\overline{c}+\overline{a}.\overline{c})=-(1+1+1-|\overline{a}+\overline{b}+\overline{c}{|}^2)$

For RHS to be minimum:

$|\overline{a}+\overline{b}+\overline{c}|=0$

$(\because RHS=|\overline{a}+\overline{b}+\overline{c}{|}^2-3)$

$\therefore \overline{a},\overline{b},\overline{c}$ must for a triange

$\therefore 2(\overline{a}.\overline{b}+\overline{b}.\overline{c}+\overline{a}.\overline{c})=-3$

Putting in (2):

$|\overline{a}-\overline{b}{|}^2+|\overline{b}-\overline{c}{|}^2+|\overline{a}-\overline{c}{|}^2=2\left(3\right)-(-3)$

$=6+3=9$

B) Answer