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 D 6

Given $\overline{a},\overline{b}$ and $\overline{c}$ are unit vector

we need maximum value, so :-

Let $\overline{a}=\hat{i}\overline{b}=\hat{j}$ and $\overline{c}=\hat{k}$

$|\overline{a}+\overline{b}{|}^2+|\overline{b}-\overline{c}{|}^2+|\overline{c}-\overline{a}{|}^2$

$|\hat{i}+\hat{j}{|}^2+|\hat{j}-\hat{k}{|}^2+|\hat{k}-\hat{i}{|}^2$

$(\sqrt{1^2+1^2}{)}^2+(\sqrt{1^2+1^2}{)}^2+(\sqrt{1^2+1^2}{)}^2$

$\Rightarrow 2+2+2=6$

$\therefore 6$ is the maximum value.

$\therefore |\overline{a}+\overline{b}|+|\overline{b}-\overline{c}|+|\overline{c}-\overline{a}{|}^2$ does not exceed $6$