Question

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

• A. $4$
• B. $9$
• C. $8$
• D. $6$

Answer 1

Answer: (B)

$9$

We have

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

$=a^2+b^2-2\overrightarrow{a}.\overrightarrow{b}+b^2+c^2-2\overrightarrow{b}.\overrightarrow{c}+a^2+c^2-2\overrightarrow{a}.\overrightarrow{c}$

Since all three vectors are unit vectors, the expression becomes $6-2(\cos \alpha +\cos \beta +\cos \gamma )$

This expression attains its maximum value when all three vectors are equally inclined with each other at ${120}^o$ and so $\cos \alpha =\cos \beta =\cos \gamma =\frac{-1}{2}$

Expression becomes $6+3=9$