Question Number 77745 by mathocean1 last updated on 09/Jan/20
![ABC is any triangle. C′ . B′ .A′ are respectively middles of [AB] . [AC] and [BC]. we suppose that AB=c AC=b BC=a. 1) u^(→ ) =a^2 BC^→ +b^(2 ) C^→ A+c^2 AB^→ is a vector Demonstrate that u^→ =(a^2 −b^2 )AC^→ +(c^2 −a^2 )AB^→ . i have done it. 2)Deduct that u^→ is not a null vector.](Q77745.png)
$$\mathrm{ABC}\:\mathrm{is}\:\mathrm{any}\:\mathrm{triangle}. \\ $$$$\mathrm{C}'\:.\:\mathrm{B}'\:\:.\mathrm{A}'\:\:\mathrm{are}\:\mathrm{respectively}\:\mathrm{middles} \\ $$$$\mathrm{of}\:\left[\mathrm{AB}\right]\:.\:\left[\mathrm{AC}\right]\:\:\mathrm{and}\:\:\left[\mathrm{BC}\right]. \\ $$$$\mathrm{we}\:\mathrm{suppose}\:\mathrm{that}\: \\ $$$$\mathrm{AB}=\mathrm{c}\:\:\:\mathrm{AC}=\mathrm{b}\:\:\:\:\mathrm{BC}=\mathrm{a}. \\ $$$$\left.\mathrm{1}\right)\:\overset{\rightarrow\:} {\mathrm{u}}=\mathrm{a}^{\mathrm{2}} \mathrm{B}\overset{\rightarrow} {\mathrm{C}}+\mathrm{b}^{\mathrm{2}\:} \overset{\rightarrow} {\mathrm{C}A}+\mathrm{c}^{\mathrm{2}} \mathrm{A}\overset{\rightarrow} {\mathrm{B}}\:\:\mathrm{is}\:\mathrm{a}\:\mathrm{vector} \\ $$$$\mathrm{Demonstrate}\:\mathrm{that}\:\overset{\rightarrow} {\mathrm{u}}=\left(\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} \right)\mathrm{A}\overset{\rightarrow} {\mathrm{C}}+\left(\mathrm{c}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} \right)\mathrm{A}\overset{\rightarrow} {\mathrm{B}}. \\ $$$${i}\:{have}\:{done}\:{it}. \\ $$$$\left.\mathrm{2}\right){D}\mathrm{educt}\:\mathrm{that}\:\overset{\rightarrow} {\mathrm{u}}\:\mathrm{is}\:\mathrm{not}\:\mathrm{a}\:\mathrm{null}\:\mathrm{vector}. \\ $$
Commented by mathocean1 last updated on 09/Jan/20
$$\mathrm{Please}\:\mathrm{can}\:\mathrm{you}\:\mathrm{help}\:\mathrm{me}... \\ $$