Question

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

• A. $4$
• B. $9$
• C. $8$
• D. $6$

Answer 1

Answer: (B)

$9$

$|\overrightarrow{a}-\overrightarrow{b}{|}^2+|\overrightarrow{a}-\overrightarrow{b}{|}^2+|\overrightarrow{a}-\overrightarrow{b}{|}^2$

$=2(a^2+b^2+c^2)-2ab\cos \theta -2b\cos \varphi -2a\cos \gamma$
Now $a=b=c=1$

Hence the above equation, reduces to

$2\left(3\right)-2(\cos \theta +\cos \varphi +\cos \gamma )$

$=2(3-(\cos \theta +\cos \varphi +\cos \gamma \left)\right)$

$=k$

Now $k$ is maximum for

$\cos \theta =\cos \varphi =\cos \gamma =-1$

However this is not possible, since the above implies

$\overrightarrow{a}$ is anit-parallel to $\overrightarrow{b}$

And $\overrightarrow{b}$ is anti-parallel to $\overrightarrow{c}$

And $\overrightarrow{c}$ is anti-parallel to $\overrightarrow{a}$ but this is not possible, since

$\overrightarrow{a}$ antiparallel to $\overrightarrow{b}$ and $\overrightarrow{b}$ antiparallel to $\overrightarrow{c}$ indicates that $\overrightarrow{a}||\overrightarrow{c}$.

Thus the above option is rules out.
Consider that all the three vectors form an equilateral triangle.
Then, $\cos \theta =\cos \varphi =\cos \gamma =\frac{-1}{2}$, or the vectors are at ${120}^0$.

Hence, we get $k$ $=2(3-\left(\frac{-3}{2}\right))$

$=2\left(\frac{9}{2}\right)$

$=9$
Thus $k_{max}=9$.

Hence, $|\overrightarrow{a}-\overrightarrow{b}{|}^2+|\overrightarrow{a}-\overrightarrow{b}{|}^2+|\overrightarrow{a}-\overrightarrow{b}{|}^2\le 9$