Question-142927

Question Number 142927 by bramlexs22 last updated on 07/Jun/21

Answered by Olaf_Thorendsen last updated on 07/Jun/21

Let (C, i^→ ,j^→ ) with C ((0),(0) ) G ((c_2 ),(0) ) = c_2 i^→ E ((0),((−c_2 )) ) = −c_2 j^→ B (((c_1 cos(−(π/2)−θ))),((c_1 sin(−(π/2)−θ))) ) = (((−c_1 sinθ)),((−c_1 cosθ)) ) D (((c_1 cos(−(π/2)−θ−(π/2)))),((c_1 sin(−(π/2)−θ−(π/2)))) ) = (((−c_1 cosθ)),((c_1 sinθ)) ) ED^(→) (((−c_1 cosθ−0)),((c_1 sinθ+c_2 )) ) = (((−c_1 cosθ)),((c_1 sinθ+c_2 )) ) BG^(→) (((c_2 +c_1 sinθ)),((0+c_1 cosθ)) ) = (((c_1 sinθ+c_2 )),((c_1 cosθ)) ) ED^(→) •BG^(→) = −c_1 cosθ(c_1 sinθ+c_2 ) +(c_1 sinθ+c_2 )(c_1 cosθ) = 0 ED^(→) •BG^(→) = 0 ⇒ ED^(→) ⊥ BG^(→)

$$\mathrm{Let}\:\left(\mathrm{C},\:\overset{\rightarrow} {{i}},\overset{\rightarrow} {{j}}\right) \\ $$$$\mathrm{with}\:\mathrm{C}\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix} \\ $$$$\mathrm{G}\begin{pmatrix}{{c}_{\mathrm{2}} }\\{\mathrm{0}}\end{pmatrix}\:=\:{c}_{\mathrm{2}} \overset{\rightarrow} {{i}} \\ $$$$\mathrm{E}\begin{pmatrix}{\mathrm{0}}\\{−{c}_{\mathrm{2}} }\end{pmatrix}\:=\:−{c}_{\mathrm{2}} \overset{\rightarrow} {{j}} \\ $$$$ \\ $$$$\mathrm{B}\begin{pmatrix}{{c}_{\mathrm{1}} \mathrm{cos}\left(−\frac{\pi}{\mathrm{2}}−\theta\right)}\\{{c}_{\mathrm{1}} \mathrm{sin}\left(−\frac{\pi}{\mathrm{2}}−\theta\right)}\end{pmatrix}\:=\:\begin{pmatrix}{−{c}_{\mathrm{1}} \mathrm{sin}\theta}\\{−\mathrm{c}_{\mathrm{1}} \mathrm{cos}\theta}\end{pmatrix} \\ $$$$\mathrm{D}\begin{pmatrix}{{c}_{\mathrm{1}} \mathrm{cos}\left(−\frac{\pi}{\mathrm{2}}−\theta−\frac{\pi}{\mathrm{2}}\right)}\\{{c}_{\mathrm{1}} \mathrm{sin}\left(−\frac{\pi}{\mathrm{2}}−\theta−\frac{\pi}{\mathrm{2}}\right)}\end{pmatrix}\:=\:\begin{pmatrix}{−{c}_{\mathrm{1}} \mathrm{cos}\theta}\\{\mathrm{c}_{\mathrm{1}} \mathrm{sin}\theta}\end{pmatrix} \\ $$$$\overset{\rightarrow} {\mathrm{ED}}\begin{pmatrix}{−{c}_{\mathrm{1}} \mathrm{cos}\theta−\mathrm{0}}\\{{c}_{\mathrm{1}} \mathrm{sin}\theta+{c}_{\mathrm{2}} }\end{pmatrix}\:=\:\begin{pmatrix}{−{c}_{\mathrm{1}} \mathrm{cos}\theta}\\{{c}_{\mathrm{1}} \mathrm{sin}\theta+{c}_{\mathrm{2}} }\end{pmatrix} \\ $$$$\overset{\rightarrow} {\mathrm{BG}}\begin{pmatrix}{{c}_{\mathrm{2}} +{c}_{\mathrm{1}} \mathrm{sin}\theta}\\{\mathrm{0}+{c}_{\mathrm{1}} \mathrm{cos}\theta}\end{pmatrix}\:=\:\begin{pmatrix}{{c}_{\mathrm{1}} \mathrm{sin}\theta+{c}_{\mathrm{2}} }\\{{c}_{\mathrm{1}} \mathrm{cos}\theta}\end{pmatrix} \\ $$$$\overset{\rightarrow} {\mathrm{ED}}\bullet\overset{\rightarrow} {\mathrm{BG}}\:=\:−{c}_{\mathrm{1}} \mathrm{cos}\theta\left({c}_{\mathrm{1}} \mathrm{sin}\theta+{c}_{\mathrm{2}} \right) \\ $$$$+\left({c}_{\mathrm{1}} \mathrm{sin}\theta+{c}_{\mathrm{2}} \right)\left({c}_{\mathrm{1}} \mathrm{cos}\theta\right)\:=\:\mathrm{0} \\ $$$$\overset{\rightarrow} {\mathrm{ED}}\bullet\overset{\rightarrow} {\mathrm{BG}}\:=\:\mathrm{0}\:\Rightarrow\:\overset{\rightarrow} {\mathrm{ED}}\:\bot\:\overset{\rightarrow} {\mathrm{BG}} \\ $$

Answered by mr W last updated on 07/Jun/21

Commented by mr W last updated on 07/Jun/21

∠BCG=∠DCE=90°+φ BC=DC CG=CE ⇒ΔBCG≡ΔDCE ⇒∠CGB=∠CED ∠CHG=∠KHE ⇒∠HKE=∠HCG=90° ⇒BG=DE & BG⊥DE

$$\angle{BCG}=\angle{DCE}=\mathrm{90}°+\phi \\ $$$${BC}={DC} \\ $$$${CG}={CE} \\ $$$$\Rightarrow\Delta{BCG}\equiv\Delta{DCE} \\ $$$$\Rightarrow\angle{CGB}=\angle{CED} \\ $$$$\angle{CHG}=\angle{KHE} \\ $$$$\Rightarrow\angle{HKE}=\angle{HCG}=\mathrm{90}° \\ $$$$\Rightarrow{BG}={DE}\:\&\:{BG}\bot{DE} \\ $$

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