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Question Number 104035 by ajfour last updated on 19/Jul/20

Find the image of point OP^(→)  = p^�   in  the line  r^� =a^� +λb^�  .

$${Find}\:{the}\:{image}\:{of}\:{point}\:\overset{\rightarrow} {{OP}}\:=\:\bar {{p}}\:\:{in} \\ $$$${the}\:{line}\:\:\bar {{r}}=\bar {{a}}+\lambda\bar {{b}}\:. \\ $$

Answered by mr W last updated on 19/Jul/20

Commented by mr W last updated on 19/Jul/20

AN=λb  (((p−a)∙b)/(∣b∣))=λ∣b∣  λ=(((p−a)∙b)/(∣b∣^2 ))  PN=a+λb−p  OQ=q=p+2PN=p+2(a+λb−p)  =2(a+λb)−p  ⇒q=2(a+(((p−a)∙b)/(∣b∣^2 ))b)−p

$${AN}=\lambda{b} \\ $$$$\frac{\left({p}−{a}\right)\centerdot{b}}{\mid{b}\mid}=\lambda\mid{b}\mid \\ $$$$\lambda=\frac{\left({p}−{a}\right)\centerdot{b}}{\mid{b}\mid^{\mathrm{2}} } \\ $$$${PN}={a}+\lambda{b}−{p} \\ $$$${OQ}={q}={p}+\mathrm{2}{PN}={p}+\mathrm{2}\left({a}+\lambda{b}−{p}\right) \\ $$$$=\mathrm{2}\left({a}+\lambda{b}\right)−{p} \\ $$$$\Rightarrow\boldsymbol{{q}}=\mathrm{2}\left(\boldsymbol{{a}}+\frac{\left(\boldsymbol{{p}}−\boldsymbol{{a}}\right)\centerdot\boldsymbol{{b}}}{\mid\boldsymbol{{b}}\mid^{\mathrm{2}} }\boldsymbol{{b}}\right)−\boldsymbol{{p}} \\ $$

Commented by ajfour last updated on 19/Jul/20

I believe this′ll be true for 3D even  Sir ?

$${I}\:{believe}\:{this}'{ll}\:{be}\:{true}\:{for}\:\mathrm{3}{D}\:{even} \\ $$$${Sir}\:? \\ $$

Commented by mr W last updated on 19/Jul/20

basically yes.

$${basically}\:{yes}. \\ $$

Commented by ajfour last updated on 19/Jul/20

((p+q)/2)=a+λb    ⇒   q=2(a+λb)−p  Now   (q−p).b=0         [(a+λb)−p].b = 0  λ=(((a−p).b)/(∣b∣^2 ))  q^� =(2a^� −p^� )+(((2(a^� −p^� ).b^� )/(∣b^� ∣^2 )))b^�

$$\frac{{p}+{q}}{\mathrm{2}}={a}+\lambda{b}\:\:\:\:\Rightarrow\:\:\:{q}=\mathrm{2}\left({a}+\lambda{b}\right)−{p} \\ $$$${Now}\:\:\:\left({q}−{p}\right).{b}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\left[\left({a}+\lambda{b}\right)−{p}\right].{b}\:=\:\mathrm{0} \\ $$$$\lambda=\frac{\left({a}−{p}\right).{b}}{\mid{b}\mid^{\mathrm{2}} } \\ $$$$\bar {{q}}=\left(\mathrm{2}\bar {{a}}−\bar {{p}}\right)+\left(\frac{\mathrm{2}\left(\bar {{a}}−\bar {{p}}\right).\bar {{b}}}{\mid\bar {{b}}\mid^{\mathrm{2}} }\right)\bar {{b}} \\ $$

Commented by mr W last updated on 19/Jul/20

yes, thanks!

$${yes},\:{thanks}! \\ $$

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