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