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

The correct option is B $9$

Given,

$\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are unit vector.

Maximum ${\left|\overrightarrow{a}-\overrightarrow{b}\right|}^2+{\left|\overrightarrow{b}-\overrightarrow{c}\right|}^2+{\left|\overrightarrow{c}-\overrightarrow{a}\right|}^2$

for maximum value all three vector will make ${120}^0$ angle between then i.e shown in figure.

$i.e,\overrightarrow{a}=\frac{\hat{i}-\hat{j}}{\sqrt{2}},\overrightarrow{b}=\frac{\hat{j}-\hat{k}}{\sqrt{2}}\&\overrightarrow{c}=\frac{\hat{k}-\hat{i}}{\sqrt{2}}$

So, maximum ${\left|\overrightarrow{a}-\overrightarrow{b}\right|}^2+{\left|\overrightarrow{b}-\overrightarrow{c}\right|}^2+{\left|\overrightarrow{c}-\overrightarrow{a}\right|}^2$

$\begin{array}{c}=\frac{1}{2}\left[{\left|\hat{i}-2\hat{j}+\hat{k}\right|}^2+{\left|\hat{j}-2\hat{k}+\hat{i}\right|}^2+{\left|\hat{k}-2\hat{i}+\hat{j}\right|}^2\right]\\ =\frac{1}{2}\left[{\left(\sqrt{1+4+1}\right)}^2+{\left(\sqrt{1+4+1}\right)}^2+{\left(\sqrt{1+4+1}\right)}^2\right]\\ =\frac{3\times 6}{2}=9\end{array}$

$\therefore$ Option $B$ is correct answer.