Question

Let $\vec{ a }, \vec{ b }, \vec{ c }$ be three vectors mutually perpendicular to each other and have same magnitude. If a vector $\vec{ r }$ satisfies.
$\vec{ a } \times\{(\vec{ r }-\vec{ b }) \times \vec{ a }\}+\vec{ b } \times\{(\vec{ r }-\vec{ c }) \times \vec{ b }\}+\vec{ c } \times\{(\vec{ r }-\vec{ a }) \times \vec{ c }\}=\vec{0}$
then $\vec{ r }$ is equal to:

• A. $\frac{1}{3}(\vec{ a }+\vec{ b }+\vec{ c })$
• B. $\frac{1}{3}(2 \vec{ a }+\vec{ b }-\vec{ c })$
• C. $\frac{1}{2}(\vec{ a }+\vec{ b }+\vec{ c })$
• D. $\frac{1}{2}(\vec{ a }+\vec{ b }+2 \vec{ c })$

Answer 1

Answer: (C)

Suppose $\vec{ r }= x \vec{ a }+ y \vec{ b }+2 \vec{ c }$
and $|\vec{ a }|=|\vec{ b }|=|\vec{ c }|= k$
$\vec{a} \times\{(\vec{r}-\vec{b}) \times \vec{a}\}+\vec{b} \times\{(\vec{r}-\vec{c}) \times \vec{b}\}+\vec{c} \times\{(\vec{r}-\vec{a}) \times \vec{c}\}=\vec{0}$
$\Rightarrow k ^{2}(\vec{r}-\vec{b})- k ^{2} \times \vec{a}+ k ^{2}(\vec{r}-\vec{c})- k ^{2} yb + k ^{2}(\overrightarrow{ r }-\overrightarrow{ a })- k ^{2} z\vec{c} =\overrightarrow{0}$
$\Rightarrow 3 \vec{r}-(\vec{a}+\vec{b}+\vec{c})-\vec{r}=\vec{0} $
$\Rightarrow \vec{r}=\frac{\vec{a}+\vec{b}+\vec{c}}{2}$