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Question Number 77745 by mathocean1 last updated on 09/Jan/20

$$\mathrm{ABC}\mathrm{is}\mathrm{any}\mathrm{triangle}.$$$${\mathrm{C}}^{\prime }.{\mathrm{B}}^{\prime }.{\mathrm{A}}^{\prime }\mathrm{are}\mathrm{respectively}\mathrm{middles}$$$$\mathrm{of}\left[\mathrm{AB}\right].\left[\mathrm{AC}\right]\mathrm{and}\left[\mathrm{BC}\right].$$$$\mathrm{we}\mathrm{suppose}\mathrm{that}$$$$\mathrm{AB}=\mathrm{c}\mathrm{AC}=\mathrm{b}\mathrm{BC}=\mathrm{a}.$$$$1)\overrightarrow{\mathrm{u}}={\mathrm{a}}^2\mathrm{B}\overrightarrow{\mathrm{C}}+{\mathrm{b}}^2\overrightarrow{\mathrm{C}A}+{\mathrm{c}}^2\mathrm{A}\overrightarrow{\mathrm{B}}\mathrm{is}\mathrm{a}\mathrm{vector}$$$$\mathrm{Demonstrate}\mathrm{that}\overrightarrow{\mathrm{u}}=({\mathrm{a}}^2-{\mathrm{b}}^2)\mathrm{A}\overrightarrow{\mathrm{C}}+({\mathrm{c}}^2-{\mathrm{a}}^2)\mathrm{A}\overrightarrow{\mathrm{B}}.$$$$ihavedoneit.$$$$2)D\mathrm{educt}\mathrm{that}\overrightarrow{\mathrm{u}}\mathrm{is}\mathrm{not}\mathrm{a}\mathrm{null}\mathrm{vector}.$$

Commented by mathocean1 last updated on 09/Jan/20

$$\mathrm{Please}\mathrm{can}\mathrm{you}\mathrm{help}\mathrm{me}...$$

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