Question Number 159606 by mr W last updated on 19/Nov/21
Commented by mr W last updated on 19/Nov/21
$${Q}\mathrm{158964} \\ $$
Answered by mr W last updated on 19/Nov/21
Commented by mr W last updated on 19/Nov/21
![METHOD I: vector method a,b,c,d are unit vectors let p=OA, q=OB OC=p(a+c) OD=q(b+d) CD=q(b+d)−p(a+c) CE=ED=(q/2)(b+d)−(p/2)(a+c) AE=AC+CE=pc+(q/2)(b+d)−(p/2)(a+c) ⇒AE=(q/2)(b+d)−(p/2)(a−c) BE=BD−ED=qd−(q/2)(b+d)+(p/2)(a+c) ⇒BE=−(q/2)(b−d)+(p/2)(a+c) AE∙BE=−(q^2 /4)(b∙b−d∙d)+((pq)/4)(a∙b−a∙d−b∙c+c∙d)+((pq)/4)(a∙b+a∙d+b∙c+c∙d)−(p^2 /4)(a∙a−c∙c) AE∙BE=−(q^2 /4)(1−1)+((pq)/4)(a∙b−a∙d−b∙c+c∙d)+((pq)/4)(a∙b+a∙d+b∙c+c∙d)−(p^2 /4)(1−1) AE∙BE=((pq)/2)(a∙b+c∙d) AE∙BE=((pq)/2)[cos θ+cos (π−θ)] AE∙BE=((pq)/2)(cos θ−cos θ) AE∙BE=0 ⇒AE⊥BE ⇒yellow angle is 90°](https://www.tinkutara.com/question/Q159614.png)
$$\mathrm{METHOD}\:\mathrm{I}:\:\mathrm{vector}\:\mathrm{method} \\ $$$$\boldsymbol{{a}},\boldsymbol{{b}},\boldsymbol{{c}},\boldsymbol{{d}}\:{are}\:{unit}\:{vectors} \\ $$$${let}\:{p}={OA},\:{q}={OB} \\ $$$$\boldsymbol{{OC}}={p}\left(\boldsymbol{{a}}+\boldsymbol{{c}}\right) \\ $$$$\boldsymbol{{OD}}={q}\left(\boldsymbol{{b}}+\boldsymbol{{d}}\right) \\ $$$$\boldsymbol{{CD}}={q}\left(\boldsymbol{{b}}+\boldsymbol{{d}}\right)−{p}\left(\boldsymbol{{a}}+\boldsymbol{{c}}\right) \\ $$$$\boldsymbol{{CE}}=\boldsymbol{{ED}}=\frac{{q}}{\mathrm{2}}\left(\boldsymbol{{b}}+\boldsymbol{{d}}\right)−\frac{{p}}{\mathrm{2}}\left(\boldsymbol{{a}}+\boldsymbol{{c}}\right) \\ $$$$\boldsymbol{{AE}}=\boldsymbol{{AC}}+\boldsymbol{{CE}}={p}\boldsymbol{{c}}+\frac{{q}}{\mathrm{2}}\left(\boldsymbol{{b}}+\boldsymbol{{d}}\right)−\frac{{p}}{\mathrm{2}}\left(\boldsymbol{{a}}+\boldsymbol{{c}}\right) \\ $$$$\Rightarrow\boldsymbol{{AE}}=\frac{{q}}{\mathrm{2}}\left(\boldsymbol{{b}}+\boldsymbol{{d}}\right)−\frac{{p}}{\mathrm{2}}\left(\boldsymbol{{a}}−\boldsymbol{{c}}\right) \\ $$$$\boldsymbol{{BE}}=\boldsymbol{{BD}}−\boldsymbol{{ED}}={q}\boldsymbol{{d}}−\frac{{q}}{\mathrm{2}}\left(\boldsymbol{{b}}+\boldsymbol{{d}}\right)+\frac{{p}}{\mathrm{2}}\left(\boldsymbol{{a}}+\boldsymbol{{c}}\right) \\ $$$$\Rightarrow\boldsymbol{{BE}}=−\frac{{q}}{\mathrm{2}}\left(\boldsymbol{{b}}−\boldsymbol{{d}}\right)+\frac{{p}}{\mathrm{2}}\left(\boldsymbol{{a}}+\boldsymbol{{c}}\right) \\ $$$$\boldsymbol{{AE}}\centerdot\boldsymbol{{BE}}=−\frac{{q}^{\mathrm{2}} }{\mathrm{4}}\left(\boldsymbol{{b}}\centerdot\boldsymbol{{b}}−\boldsymbol{{d}}\centerdot\boldsymbol{{d}}\right)+\frac{{pq}}{\mathrm{4}}\left(\boldsymbol{{a}}\centerdot\boldsymbol{{b}}−\boldsymbol{{a}}\centerdot\boldsymbol{{d}}−\boldsymbol{{b}}\centerdot\boldsymbol{{c}}+\boldsymbol{{c}}\centerdot\boldsymbol{{d}}\right)+\frac{{pq}}{\mathrm{4}}\left(\boldsymbol{{a}}\centerdot\boldsymbol{{b}}+\boldsymbol{{a}}\centerdot\boldsymbol{{d}}+\boldsymbol{{b}}\centerdot\boldsymbol{{c}}+\boldsymbol{{c}}\centerdot\boldsymbol{{d}}\right)−\frac{{p}^{\mathrm{2}} }{\mathrm{4}}\left(\boldsymbol{{a}}\centerdot\boldsymbol{{a}}−\boldsymbol{{c}}\centerdot\boldsymbol{{c}}\right) \\ $$$$\boldsymbol{{AE}}\centerdot\boldsymbol{{BE}}=−\frac{{q}^{\mathrm{2}} }{\mathrm{4}}\left(\mathrm{1}−\mathrm{1}\right)+\frac{{pq}}{\mathrm{4}}\left(\boldsymbol{{a}}\centerdot\boldsymbol{{b}}−\boldsymbol{{a}}\centerdot\boldsymbol{{d}}−\boldsymbol{{b}}\centerdot\boldsymbol{{c}}+\boldsymbol{{c}}\centerdot\boldsymbol{{d}}\right)+\frac{{pq}}{\mathrm{4}}\left(\boldsymbol{{a}}\centerdot\boldsymbol{{b}}+\boldsymbol{{a}}\centerdot\boldsymbol{{d}}+\boldsymbol{{b}}\centerdot\boldsymbol{{c}}+\boldsymbol{{c}}\centerdot\boldsymbol{{d}}\right)−\frac{{p}^{\mathrm{2}} }{\mathrm{4}}\left(\mathrm{1}−\mathrm{1}\right) \\ $$$$\boldsymbol{{AE}}\centerdot\boldsymbol{{BE}}=\frac{{pq}}{\mathrm{2}}\left(\boldsymbol{{a}}\centerdot\boldsymbol{{b}}+\boldsymbol{{c}}\centerdot\boldsymbol{{d}}\right) \\ $$$$\boldsymbol{{AE}}\centerdot\boldsymbol{{BE}}=\frac{{pq}}{\mathrm{2}}\left[\mathrm{cos}\:\theta+\mathrm{cos}\:\left(\pi−\theta\right)\right] \\ $$$$\boldsymbol{{AE}}\centerdot\boldsymbol{{BE}}=\frac{{pq}}{\mathrm{2}}\left(\mathrm{cos}\:\theta−\mathrm{cos}\:\theta\right) \\ $$$$\boldsymbol{{AE}}\centerdot\boldsymbol{{BE}}=\mathrm{0} \\ $$$$\Rightarrow\boldsymbol{{AE}}\bot\boldsymbol{{BE}} \\ $$$$\Rightarrow{yellow}\:{angle}\:{is}\:\mathrm{90}° \\ $$
Commented by Tawa11 last updated on 28/Nov/21
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Answered by mr W last updated on 19/Nov/21
