Let-p-q-r-be-three-mutually-perpendicular-vectors-of-the-same-magnitude-If-a-vector-x-satisfies-the-equation-p-x-q-p-q-x-r-q-r-x-p-r-0-then-x-is-given-by-

Question Number 19262 by djatmiko last updated on 08/Aug/17

Let \boldsymbol p, \boldsymbol q, \boldsymbol r be three mutually perpendicular

vectors of the same magnitude . If a vector

\boldsymbol x satisfies the equation

\boldsymbol p \times {\boldsymbol x - \boldsymbol q) \times \boldsymbol p} + \boldsymbol q \times {\boldsymbol x - \boldsymbol r) \times \boldsymbol q}

+ \boldsymbol r \times {\boldsymbol x - \boldsymbol p) \times \boldsymbol r} = 0, then \boldsymbol x is

given by

Answered by ajfour last updated on 08/Aug/17

= (\bar{p} . \bar{p}) (\bar{x} - \bar{q}) - [\bar{p} . (\bar{x} - \bar{q})] \bar{p} +

(\bar{q} . \bar{q}) (\bar{x} - \bar{r}) - [\bar{q} . (\bar{x} - \bar{r})] \bar{q} +

(\bar{r} . \bar{r}) (\bar{x} - \bar{p}) - [\bar{r} . (\bar{x} - \bar{p})] \bar{r}

= (p^{2} + q^{2} + r^{2}) \bar{x} - [(\bar{p} . \bar{x}) \bar{p} + (\bar{q} . \bar{x}) \bar{q} + (\bar{r} . \bar{x}) \bar{r}]

- (p^{2} + q^{2} + r^{2}) (\bar{p} + \bar{q} + \bar{r}) +

(\bar{p} . \bar{q}) \bar{p} + (\bar{q} . \bar{r}) \bar{q} + (\bar{r} . \bar{p}) \bar{r} = 0

\bar{p}, \bar{q}, and \bar{r} are mutually ⊥,

and of equal magnitudes; let

their manitudes be a .

\Rightarrow {3 a}^{2} [\bar{x} - (\bar{p} + \bar{q} + \bar{r})] = a (x_{p} \bar{p} + x_{q} \bar{q} + x_{r} \bar{r})

\Rightarrow {3 a}^{2} [\bar{x} - (\bar{p} + \bar{q} + \bar{r})] = a^{2} (\bar{x})

\Rightarrow \bar{x} = \frac{3}{2} (\bar{p} + \bar{q} + \bar{r}) .

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