Question Number 19262 by djatmiko last updated on 08/Aug/17
$$\mathrm{Let}\:\boldsymbol{\mathrm{p}},\:\boldsymbol{\mathrm{q}},\:\boldsymbol{\mathrm{r}}\:\mathrm{be}\:\mathrm{three}\:\mathrm{mutually}\:\mathrm{perpendicular} \\ $$$$\mathrm{vectors}\:\mathrm{of}\:\mathrm{the}\:\mathrm{same}\:\mathrm{magnitude}.\:\mathrm{If}\:\mathrm{a}\:\mathrm{vector} \\ $$$$\boldsymbol{\mathrm{x}}\:\:\mathrm{satisfies}\:\mathrm{the}\:\mathrm{equation}\: \\ $$$$\left.\boldsymbol{\mathrm{p}}\left.\:×\left\{\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{q}}\right)×\boldsymbol{\mathrm{p}}\right\}+\boldsymbol{\mathrm{q}}×\left\{\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{r}}\right)×\boldsymbol{\mathrm{q}}\right\} \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\:\:\boldsymbol{\mathrm{r}}×\left\{\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{p}}\right)×\boldsymbol{\mathrm{r}}\right\}=\mathrm{0},\:\mathrm{then}\:\boldsymbol{\mathrm{x}}\:\mathrm{is} \\ $$$$\mathrm{given}\:\mathrm{by} \\ $$
Answered by ajfour last updated on 08/Aug/17
![= (p^ .p^ )(x^ −q^ )−[p^ .(x^ −q^ )]p^ + (q^ .q^ )(x^ −r^ )−[q^ .(x^ −r^ )]q^ + (r^ .r^ )(x^ −p^ )−[r^ .(x^ −p^ )]r^ = (p^2 +q^2 +r^2 )x^ −[(p^ .x^ )p^ +(q^ .x^ )q^ +(r^ .x^ )r^ ] −(p^2 +q^2 +r^2 )(p^ +q^ +r^ )+ (p^ .q^ )p^ +(q^ .r^ )q^ +(r^ .p^ )r^ =0 p^ , q^ , and r^ are mutually ⊥, and of equal magnitudes ;let their manitudes be a. ⇒3a^2 [x^ −(p^ +q^ +r^ )]=a(x_p p^ +x_q q^ +x_r r^ ) ⇒ 3a^2 [x^ −(p^ +q^ +r^ )]=a^2 (x^ ) ⇒ x^ =(3/2)(p^ +q^ +r^ ) .](Q19263.png)
$$=\:\left(\bar {\mathrm{p}}.\bar {\mathrm{p}}\right)\left(\bar {\mathrm{x}}−\bar {\mathrm{q}}\right)−\left[\bar {\mathrm{p}}.\left(\bar {\mathrm{x}}−\bar {\mathrm{q}}\right)\right]\bar {\mathrm{p}}+ \\ $$$$\:\:\:\:\:\left(\bar {\mathrm{q}}.\bar {\mathrm{q}}\right)\left(\bar {\mathrm{x}}−\bar {\mathrm{r}}\right)−\left[\bar {\mathrm{q}}.\left(\bar {\mathrm{x}}−\bar {\mathrm{r}}\right)\right]\bar {\mathrm{q}}+ \\ $$$$\:\:\:\:\:\left(\bar {\mathrm{r}}.\bar {\mathrm{r}}\right)\left(\bar {\mathrm{x}}−\bar {\mathrm{p}}\right)−\left[\bar {\mathrm{r}}.\left(\bar {\mathrm{x}}−\bar {\mathrm{p}}\right)\right]\bar {\mathrm{r}} \\ $$$$=\:\left(\mathrm{p}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} \right)\bar {\mathrm{x}}−\left[\left(\bar {\mathrm{p}}.\bar {\mathrm{x}}\right)\bar {\mathrm{p}}+\left(\bar {\mathrm{q}}.\bar {\mathrm{x}}\right)\bar {\mathrm{q}}+\left(\bar {\mathrm{r}}.\bar {\mathrm{x}}\right)\bar {\mathrm{r}}\right] \\ $$$$\:\:\:\:\:−\left(\mathrm{p}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} \right)\left(\bar {\mathrm{p}}+\bar {\mathrm{q}}+\bar {\mathrm{r}}\right)+ \\ $$$$\:\:\:\:\:\:\left(\bar {\mathrm{p}}.\bar {\mathrm{q}}\right)\bar {\mathrm{p}}+\left(\bar {\mathrm{q}}.\bar {\mathrm{r}}\right)\bar {\mathrm{q}}+\left(\bar {\mathrm{r}}.\bar {\mathrm{p}}\right)\bar {\mathrm{r}}=\mathrm{0} \\ $$$$\:\bar {\mathrm{p}},\:\bar {\mathrm{q}},\:\mathrm{and}\:\bar {\mathrm{r}}\:\mathrm{are}\:\mathrm{mutually}\:\bot, \\ $$$$\mathrm{and}\:\mathrm{of}\:\mathrm{equal}\:\mathrm{magnitudes}\:;\mathrm{let} \\ $$$$\mathrm{their}\:\mathrm{manitudes}\:\mathrm{be}\:\mathrm{a}. \\ $$$$\Rightarrow\mathrm{3a}^{\mathrm{2}} \left[\bar {\mathrm{x}}−\left(\bar {\mathrm{p}}+\bar {\mathrm{q}}+\bar {\mathrm{r}}\right)\right]=\mathrm{a}\left(\mathrm{x}_{\mathrm{p}} \bar {\mathrm{p}}+\mathrm{x}_{\mathrm{q}} \bar {\mathrm{q}}+\mathrm{x}_{\mathrm{r}} \bar {\mathrm{r}}\right) \\ $$$$\Rightarrow\:\mathrm{3a}^{\mathrm{2}} \left[\bar {\mathrm{x}}−\left(\bar {\mathrm{p}}+\bar {\mathrm{q}}+\bar {\mathrm{r}}\right)\right]=\mathrm{a}^{\mathrm{2}} \left(\bar {\mathrm{x}}\right) \\ $$$$\Rightarrow\:\:\bar {\mathrm{x}}=\frac{\mathrm{3}}{\mathrm{2}}\left(\bar {\mathrm{p}}+\bar {\mathrm{q}}+\bar {\mathrm{r}}\right)\:. \\ $$