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

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

Answered by ajfour last updated on 08/Aug/17

$$=(\overline{\mathrm{p}}.\overline{\mathrm{p}})(\overline{\mathrm{x}}-\overline{\mathrm{q}})-[\overline{\mathrm{p}}.(\overline{\mathrm{x}}-\overline{\mathrm{q}})]\overline{\mathrm{p}}+$$$$(\overline{\mathrm{q}}.\overline{\mathrm{q}})(\overline{\mathrm{x}}-\overline{\mathrm{r}})-[\overline{\mathrm{q}}.(\overline{\mathrm{x}}-\overline{\mathrm{r}})]\overline{\mathrm{q}}+$$$$(\overline{\mathrm{r}}.\overline{\mathrm{r}})(\overline{\mathrm{x}}-\overline{\mathrm{p}})-[\overline{\mathrm{r}}.(\overline{\mathrm{x}}-\overline{\mathrm{p}})]\overline{\mathrm{r}}$$$$=({\mathrm{p}}^2+{\mathrm{q}}^2+{\mathrm{r}}^2)\overline{\mathrm{x}}-[(\overline{\mathrm{p}}.\overline{\mathrm{x}})\overline{\mathrm{p}}+(\overline{\mathrm{q}}.\overline{\mathrm{x}})\overline{\mathrm{q}}+(\overline{\mathrm{r}}.\overline{\mathrm{x}})\overline{\mathrm{r}}]$$$$-({\mathrm{p}}^2+{\mathrm{q}}^2+{\mathrm{r}}^2)(\overline{\mathrm{p}}+\overline{\mathrm{q}}+\overline{\mathrm{r}})+$$$$(\overline{\mathrm{p}}.\overline{\mathrm{q}})\overline{\mathrm{p}}+(\overline{\mathrm{q}}.\overline{\mathrm{r}})\overline{\mathrm{q}}+(\overline{\mathrm{r}}.\overline{\mathrm{p}})\overline{\mathrm{r}}=0$$$$\overline{\mathrm{p}},\overline{\mathrm{q}},\mathrm{and}\overline{\mathrm{r}}\mathrm{are}\mathrm{mutually}\mathrm{\perp },$$$$\mathrm{and}\mathrm{of}\mathrm{equal}\mathrm{magnitudes};\mathrm{let}$$$$\mathrm{their}\mathrm{manitudes}\mathrm{be}\mathrm{a}.$$$$\Rightarrow {3\mathrm{a}}^2[\overline{\mathrm{x}}-(\overline{\mathrm{p}}+\overline{\mathrm{q}}+\overline{\mathrm{r}})]=\mathrm{a}({\mathrm{x}}_{\mathrm{p}}\overline{\mathrm{p}}+{\mathrm{x}}_{\mathrm{q}}\overline{\mathrm{q}}+{\mathrm{x}}_{\mathrm{r}}\overline{\mathrm{r}})$$$$\Rightarrow {3\mathrm{a}}^2[\overline{\mathrm{x}}-(\overline{\mathrm{p}}+\overline{\mathrm{q}}+\overline{\mathrm{r}})]={\mathrm{a}}^2\left(\overline{\mathrm{x}}\right)$$$$\Rightarrow \overline{\mathrm{x}}=\frac{3}{2}(\overline{\mathrm{p}}+\overline{\mathrm{q}}+\overline{\mathrm{r}}).$$

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