Question-59407

Question Number 59407 by cesar.marval.larez@gmail.com last updated on 09/May/19

Commented by cesar.marval.larez@gmail.com last updated on 09/May/19

$$\mathrm{I}\:\mathrm{need}\:\mathrm{help}\:\mathrm{please} \\ $$

Answered by MJS last updated on 10/May/19

A′=A+2AP^(→) =A+2(P−A)=2P−A= =2 (((−2)),(1),(0) ) − ((2),(1),(3) ) = (((−6)),(1),((−3)) ) r= { ((x=1+2t)),((y=0)),((z=−2−t)) :}= (((1+2t)),(0),((−2−t)) ) r′=2P−r=2 (((−2)),(1),(0) ) − (((1+2t)),(0),((−2−t)) ) = (((−5−2t)),(2),((2+t)) ) = = { ((x=−5−2t)),((y=2)),((z=2+t)) :} α: x−11y+2z+3=0 x′=−4−x y′=2−y z′=0−z ⇒ α′: (−4−x)−11(2−y)+2(−z)+3=0 ⇒ α′: −x+11y−2z−23=0

$${A}'={A}+\mathrm{2}\overset{\rightarrow} {{AP}}={A}+\mathrm{2}\left({P}−{A}\right)=\mathrm{2}{P}−{A}= \\ $$$$=\mathrm{2}\begin{pmatrix}{−\mathrm{2}}\\{\mathrm{1}}\\{\mathrm{0}}\end{pmatrix}\:−\begin{pmatrix}{\mathrm{2}}\\{\mathrm{1}}\\{\mathrm{3}}\end{pmatrix}\:=\begin{pmatrix}{−\mathrm{6}}\\{\mathrm{1}}\\{−\mathrm{3}}\end{pmatrix} \\ $$$$ \\ $$$${r}=\begin{cases}{{x}=\mathrm{1}+\mathrm{2}{t}}\\{{y}=\mathrm{0}}\\{{z}=−\mathrm{2}−{t}}\end{cases}=\begin{pmatrix}{\mathrm{1}+\mathrm{2}{t}}\\{\mathrm{0}}\\{−\mathrm{2}−{t}}\end{pmatrix} \\ $$$${r}'=\mathrm{2}{P}−{r}=\mathrm{2}\begin{pmatrix}{−\mathrm{2}}\\{\mathrm{1}}\\{\mathrm{0}}\end{pmatrix}\:−\begin{pmatrix}{\mathrm{1}+\mathrm{2}{t}}\\{\mathrm{0}}\\{−\mathrm{2}−{t}}\end{pmatrix}\:=\begin{pmatrix}{−\mathrm{5}−\mathrm{2}{t}}\\{\mathrm{2}}\\{\mathrm{2}+{t}}\end{pmatrix}\:= \\ $$$$=\begin{cases}{{x}=−\mathrm{5}−\mathrm{2}{t}}\\{{y}=\mathrm{2}}\\{{z}=\mathrm{2}+{t}}\end{cases} \\ $$$$ \\ $$$$\alpha:\:{x}−\mathrm{11}{y}+\mathrm{2}{z}+\mathrm{3}=\mathrm{0} \\ $$$${x}'=−\mathrm{4}−{x} \\ $$$${y}'=\mathrm{2}−{y} \\ $$$${z}'=\mathrm{0}−{z} \\ $$$$\Rightarrow\:\alpha':\:\left(−\mathrm{4}−{x}\right)−\mathrm{11}\left(\mathrm{2}−{y}\right)+\mathrm{2}\left(−{z}\right)+\mathrm{3}=\mathrm{0} \\ $$$$\Rightarrow\:\alpha':\:−{x}+\mathrm{11}{y}−\mathrm{2}{z}−\mathrm{23}=\mathrm{0} \\ $$