Question

If $\mathit{a}\mathbf{,}\mathbf{}\mathit{b}and\mathit{c}$ are unit vectors such that$\mathit{a}+\mathit{b}+\mathit{c}=0$, find the value of vector $\mathit{a}\mathbf{.}\mathit{b}+\mathit{b}\mathbf{.}\mathit{c}+\mathit{c}\mathbf{.}\mathit{a}?$

• A. $\frac{1}{2}$
• B. $0$
• C. $1$
• D. $\frac{-3}{2}$

Answer 1

The correct option is D

$\frac{-3}{2}$

Explanation for correct option:

Finding the value of vector:

Given: $a,bandc$ are unit vectors such that$a+b+c=0$,

$$\begin{gathered} (a+b+c).(a+b+c)={\left|a\right|}^2+{\left|b\right|}^2+{\left|c\right|}^2+2(a.b+b.c+c.a) \\ {\left|a+b+c\right|}^2=1+1+1+2(a.b+b.c+c.a) \\ 0=3+2(a.b+b.c+c.a) \\ \frac{-3}{2}=a.b+b.c+c.a \\ a.b+b.c+c.a=\frac{-3}{2} \end{gathered}$$

Hence, the option$\mathbf{\left(}\mathit{D}\mathbf{\right)}$ is correct.