Prove the following

Question:

Let $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}$ be three vectors mutually perpendicular to each other and have same magnitude. If a vector $\overrightarrow{\mathrm{r}}$ satisfies.

$\overrightarrow{\mathrm{a}} \times\{(\overrightarrow{\mathrm{r}}-\overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{a}}\}+\overrightarrow{\mathrm{b}} \times\{(\overrightarrow{\mathrm{r}}-\overrightarrow{\mathrm{c}}) \times \overrightarrow{\mathrm{b}}\}+\overrightarrow{\mathrm{c}} \times\{(\overrightarrow{\mathrm{r}}-\overrightarrow{\mathrm{a}}) \times \overrightarrow{\mathrm{c}}\}=\overrightarrow{0}$

then $\overrightarrow{\mathrm{r}}$ is equal to :

1. $\frac{1}{3}(\vec{a}+\vec{b}+\vec{c})$

2. $\frac{1}{3}(2 \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}})$

3. $\frac{1}{2}(\vec{a}+\vec{b}+\vec{c})$

4. $\frac{1}{2}(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+2 \overrightarrow{\mathrm{c}})$

Correct Option: , 3

Solution:

Suppose $\overrightarrow{\mathrm{r}}=\mathrm{x} \overrightarrow{\mathrm{a}}+\mathrm{yb}+2 \overrightarrow{\mathrm{c}}$

and $|\overrightarrow{\mathrm{a}}|=|\overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{c}}|=\mathrm{k}$

$\overrightarrow{\mathrm{a}} \times\{(\overrightarrow{\mathrm{r}}-\overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{a}}\}+\overrightarrow{\mathrm{b}} \times\{(\overrightarrow{\mathrm{r}}-\overrightarrow{\mathrm{c}}) \times \overrightarrow{\mathrm{b}}\}+\overrightarrow{\mathrm{c}} \times\{(\overrightarrow{\mathrm{r}}-\overrightarrow{\mathrm{a}}) \times \overrightarrow{\mathrm{c}}\}=\overrightarrow{0}$

$\Rightarrow \mathrm{k}^{2}(\overrightarrow{\mathrm{r}}-\overrightarrow{\mathrm{b}})-\mathrm{k}^{2} \mathrm{x} \overrightarrow{\mathrm{a}}+\mathrm{k}^{2}(\overrightarrow{\mathrm{r}}-\overrightarrow{\mathrm{c}})-\mathrm{k}^{2} \mathrm{y} \overrightarrow{\mathrm{b}}+$$\mathrm{k}^{2}(\overrightarrow{\mathrm{r}}-\overrightarrow{\mathrm{a}})-\mathrm{k}^{2} \mathrm{z} \overrightarrow{\mathrm{c}}=\overrightarrow{0}$

$\Rightarrow 3 \overrightarrow{\mathrm{r}}-(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})-\overrightarrow{\mathrm{r}}=\overrightarrow{0}$

$\Rightarrow \overrightarrow{\mathrm{r}}=\frac{\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}}{2}$