Question

If $\vec{a},\vec{b}$ and $\vec{c}$ are unit vectors, such that $\vec{a}+\vec{b}+\vec{c}=0$ then $3\vec{a}.\vec{b}+\overrightarrow{2b}.\vec{c}+\vec{c}.\vec{a}$

• A. -3
• B. 3
• C. -1
• D. 1

Answer 1

Answer: (A)

$\vec{a},\vec{b},\vec{c}$ are unit vectors, then $|\vec{a}|=|\vec{b}|=|\vec{c}|=1$
Given, $\vec{a}+\vec{b}+\vec{c}=0$
$\vec{a}=-(\vec{b}+\vec{c})$
Squaring on both sides
$a^{\vec{2}}=(\vec{b}+\vec{c}{)}^2$
$a^{\vec{2}}={\vec{b}}^2+c^{\vec{2}}+2\vec{b}\cdot \vec{c}$
$|\vec{a}{|}^2=|\vec{b}{|}^2+|\vec{c}{|}^2+2(\vec{b}\cdot \vec{c})$
$\left(\because a^{\vec{2}}=|\vec{a}{|}^2\right)$
$1=1+1+2(\vec{b}\cdot \vec{c})$
$\Rightarrow \vec{b}\cdot \vec{c}=-1/2$
Similarly, $\vec{a}\cdot \vec{b}=\vec{c}\cdot \vec{a}=-1/2$
Hence, $3a\cdot \vec{b}+2\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a}$
$=3(-1/2)+2(-1/2)+(-1/2)$
$=(3+2+1)(-1/2)$
$=6(-1/2)=-3$