1. Tardigrade
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  4. Let p , q , r be three mutually perpendicular vectors of the same magnitude. If a vector vecx satisfies the equation p ×[( x - q ) × p ] + q ×[( x - r ) × q ] + r ×[( x - p ) × r ]= 0 , then x is given by

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Q. Let $\overrightarrow{ p }, \overrightarrow{ q }, \overrightarrow{ r }$ be three mutually perpendicular vectors of the same magnitude. If a vector $\vec{x}$ 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

IIT JEEIIT JEE 1997Vector Algebra

Solution:

Since, $\overrightarrow{ p }, \overrightarrow{ q }, \overrightarrow{ r }$ are mutually perpendicular vectors of same magnitude, so let us consider
$|\overrightarrow{ p }|=|\overrightarrow{ q }|=|\overrightarrow{ r }|=\lambda $ and
$\overrightarrow{ p } \cdot \overrightarrow{ q }=\overrightarrow{ q } \cdot \overrightarrow{ r }=\overrightarrow{ r } \cdot \overrightarrow{ p }=0 .....$ (i)
Given, $\overrightarrow{ p } \times\{(\overrightarrow{ x }-\overrightarrow{ q }) \times \overrightarrow{ p }\}+\overrightarrow{ q } \times\{(\overrightarrow{ x }-\overrightarrow{ r }) \times \overrightarrow{ q }\}$
$+\overrightarrow{ r } \times\{(\overrightarrow{ x }-\overrightarrow{ p }) \times \overrightarrow{ r }\}=\overrightarrow{0}$
$\Rightarrow (\overrightarrow{ p } \cdot \overrightarrow{ p })(\overrightarrow{ x }-\overrightarrow{ q })-\{\overrightarrow{ p } \cdot(\overrightarrow{ x }-\overrightarrow{ q })\} \overrightarrow{ p }+(\overrightarrow{ q } \cdot \overrightarrow{ q })(\overrightarrow{ x }-\overrightarrow{ r })$
$-\{\overrightarrow{ q } \cdot(\overrightarrow{ x }-\overrightarrow{ r })\} \overrightarrow{ q }+(\overrightarrow{ r } \cdot \overrightarrow{ r })(\overrightarrow{ x }-\overrightarrow{ p })-\{\overrightarrow{ r } \cdot(\overrightarrow{ x }-\overrightarrow{ p })\} \overrightarrow{ r }=0 $
$\Rightarrow \overrightarrow{ x }(\overrightarrow{ p } \cdot \overrightarrow{ p })+(\overrightarrow{ q } \cdot \overrightarrow{ q })+(\overrightarrow{ r } \cdot \overrightarrow{ r })\}-(\overrightarrow{ p } \cdot \overrightarrow{ p }) \overrightarrow{ q }$
$-(\overrightarrow{ q } \cdot \overrightarrow{ q }) r-(\overrightarrow{ r } \cdot \overrightarrow{ r }) \overrightarrow{ p }=(\overrightarrow{ x } \cdot \overrightarrow{ p }) \overrightarrow{ p }+(\overrightarrow{ x } \cdot \overrightarrow{ q }) \overrightarrow{ q }+(\overrightarrow{ x } \cdot \overrightarrow{ r }) \overrightarrow{ r }$
$\Rightarrow 3 \overrightarrow{ x }|\lambda|^{2}-(\overrightarrow{ p }+\overrightarrow{ q }+\overrightarrow{ r })|\lambda|^{2}=(\overrightarrow{ x } \cdot \overrightarrow{ p }) \overrightarrow{ p }$
$+(\overrightarrow{ x } \cdot \overrightarrow{ q }) \overrightarrow{ q }+(\overrightarrow{ x } \cdot \overrightarrow{ r }) \overrightarrow{ r } ....$(ii)
Taking dot of Eq. (ii) with $\overrightarrow{ p }$, we get
$3(\overrightarrow{ x } \cdot \overrightarrow{ p })|\lambda|^{2}-|\lambda|^{4}=(\overrightarrow{ x } \cdot \overrightarrow{ p })|\lambda|^{2} \Rightarrow \overrightarrow{ x } \cdot \overrightarrow{ p }=\frac{1}{2}|\lambda|^{2}$
Similarly, taking dot of Eq. (ii) with $\overrightarrow{ q }$ and $\overrightarrow{ r }$ respectively, we get
$\overrightarrow{ x } \cdot \overrightarrow{ q }=\frac{|\lambda|^{2}}{2}=\overrightarrow{ x } \cdot \overrightarrow{ r }$
$\therefore$ Eq. (ii) becomes
$3 \overrightarrow{ x }|\lambda|^{2}-(\overrightarrow{ p }+\overrightarrow{ q }+\overrightarrow{ r })|\lambda|^{2}=\frac{|\lambda|^{2}}{2}(\overrightarrow{ p }+\overrightarrow{ q }+\overrightarrow{ r })$
$\Rightarrow 3 \overrightarrow{ x }=\frac{1}{2}(\overrightarrow{ p }+\overrightarrow{ q }+\overrightarrow{ r })+(\overrightarrow{ p }+\overrightarrow{ q }+\overrightarrow{ r }) \Rightarrow \vec{x}=\frac{1}{2}(\overrightarrow{ p }+\overrightarrow{ q }+\overrightarrow{ r })$