Question

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

• A. -3
• B. 3
• C. -1
• D. 1

Answer 1

Answer: (A)

-3

Answer 2

$\vec{ a }, \vec{ b }, \vec{ c }$ 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, $3 a \cdot \vec{ b }+2 \vec{ b } \cdot \vec{ c }+\vec{ c } \cdot \vec{ a }$
$=3(-1 / 2)+2(-1 / 2)+(-1 / 2)$
$=(3+2+1)(-1 / 2)$
$=6(-1 / 2)=-3$