Question

Let $\overrightarrow{p},\overrightarrow{q},\overrightarrow{r}$ be three mutually perpendicular vectors of the same magnitude. If a vectors $\overrightarrow{x}$ satisfies the equation $\overrightarrow{p}\times \{(\overrightarrow{x}-\overrightarrow{q})\times \overrightarrow{p}\}+\overrightarrow{q}\times \{(\overrightarrow{x}-\overrightarrow{r})\times \overrightarrow{q}\}+\overrightarrow{r}\times \{(\overrightarrow{x}-\overrightarrow{p})\times \overrightarrow{r}\}=\overrightarrow{0}$ then $\overrightarrow{x}$ is given by

• A. $\frac{1}{2}(\overrightarrow{p}+\overrightarrow{q}-2\overrightarrow{r})$
• B. $\frac{1}{2}(\overrightarrow{p}+\overrightarrow{q}+\overrightarrow{r})$
• C. $\frac{1}{3}(\overrightarrow{p}+\overrightarrow{q}+\overrightarrow{r})$
• D. $\frac{1}{3}(2\overrightarrow{p}+\overrightarrow{q}+\overrightarrow{r})$

Answer 1

The correct option is B $\frac{1}{2}(\overrightarrow{p}+\overrightarrow{q}+\overrightarrow{r})$

$\overrightarrow{p}\times \{(\overrightarrow{x}-\overrightarrow{q})\times \overrightarrow{p}\}+\overrightarrow{q}\times \{(\overrightarrow{x}-\overrightarrow{r})\times \overrightarrow{q}\}+\overrightarrow{r}\times \{(\overrightarrow{x}-\overrightarrow{p})\times \overrightarrow{r}\}=\overrightarrow{0}$

$\Rightarrow (\overrightarrow{p}.\overrightarrow{p})(\overrightarrow{x}-\overrightarrow{q})-\left[\right(\overrightarrow{x}-\overrightarrow{q}).\overrightarrow{p}]\overrightarrow{p}+(\overrightarrow{q}.\overrightarrow{q})(\overrightarrow{x}-\overrightarrow{r})-\left[\right(\overrightarrow{x}-\overrightarrow{r}).\overrightarrow{q}]\overrightarrow{q}+(\overrightarrow{r}.\overrightarrow{r})(\overrightarrow{x}-\overrightarrow{p})-\left[\right(\overrightarrow{x}-\overrightarrow{p}).\overrightarrow{r}]\overrightarrow{r}=\overrightarrow{0}$

$\Rightarrow |\overrightarrow{p}{|}^2(\overrightarrow{x}-\overrightarrow{q})-(\overrightarrow{p}.\overrightarrow{x})\overrightarrow{p}+|\overrightarrow{q}{|}^2(\overrightarrow{x}-\overrightarrow{r})-(\overrightarrow{q}.\overrightarrow{x})\overrightarrow{q}+|\overrightarrow{r}{|}^2(\overrightarrow{x}-\overrightarrow{p})-(\overrightarrow{r}.\overrightarrow{x})\overrightarrow{r}=\overrightarrow{0}$

Simplifying, we have $|\overrightarrow{p}{|}^2(3\overrightarrow{x}-\overrightarrow{p}-\overrightarrow{q}-\overrightarrow{r})=(\overrightarrow{p}.\overrightarrow{x})\overrightarrow{p}+(\overrightarrow{q}.\overrightarrow{x})\overrightarrow{q}+(\overrightarrow{r}.\overrightarrow{x})\overrightarrow{r}$

Lets consider the expression written on $RHS$:

$(\overrightarrow{p}.\overrightarrow{x})\overrightarrow{p}+(\overrightarrow{q}.\overrightarrow{x})\overrightarrow{q}+(\overrightarrow{r}.\overrightarrow{x})\overrightarrow{r}$

Let $\overrightarrow{x}=a\overrightarrow{p}+b\overrightarrow{q}+\overrightarrow{r}$

$\overrightarrow{x}.\overrightarrow{p}=a|\overrightarrow{p}{|}^2$

$\overrightarrow{x}.\overrightarrow{q}=b|\overrightarrow{q}{|}^2=b|\overrightarrow{p}{|}^2$

$\overrightarrow{x}.\overrightarrow{r}=c|\overrightarrow{r}{|}^2=c|\overrightarrow{p}{|}^2$

So the expression RHS can be reduced to

RHS $=|\overrightarrow{p}{|}^2(a\overrightarrow{p}+b\overrightarrow{q}+c\overrightarrow{r})$

$\Rightarrow$ RHS $=|\overrightarrow{p}{|}^2\overrightarrow{x}$

Hence the whole equation reduces to:

$|\overrightarrow{p}{|}^2(3\overrightarrow{x}-\overrightarrow{p}-\overrightarrow{q}-\overrightarrow{r})=|\overrightarrow{p}{|}^2\overrightarrow{x}$

Thus, we have $\overrightarrow{x}$ as $\frac{1}{2}(\overrightarrow{p}+\overrightarrow{q}+\overrightarrow{r})$

Hence, option $B$.