Question Number 95779 by i jagooll last updated on 27/May/20
$$\mathrm{if}\:\mathrm{plane}\:\mathrm{3x}+\mathrm{4y}+\mathrm{tz}=\mathrm{2}\:\mathrm{and}\: \\ $$$$\mathrm{kx}+\mathrm{6y}+\mathrm{5z}−\mathrm{2}=\mathrm{0}\:\mathrm{are}\:\mathrm{parallel}. \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{k}\:\mathrm{and}\:\mathrm{t}\: \\ $$
Answered by john santu last updated on 27/May/20
$$\frac{\mathrm{3}}{\mathrm{k}}\:=\:\frac{\mathrm{4}}{\mathrm{6}}\:=\:\frac{\mathrm{t}}{\mathrm{5}} \\ $$$$\begin{cases}{\mathrm{k}=\frac{\mathrm{18}}{\mathrm{4}}=\frac{\mathrm{9}}{\mathrm{2}}\:}\\{\mathrm{t}\:=\:\frac{\mathrm{20}}{\mathrm{6}}=\frac{\mathrm{10}}{\mathrm{3}}}\end{cases} \\ $$$$ \\ $$
Commented by i jagooll last updated on 27/May/20
$$\mathrm{thank}\:\mathrm{you} \\ $$
Answered by Rio Michael last updated on 28/May/20
$$\mathrm{normal}\:\mathrm{for}\:\mathrm{plane}\:\mathrm{1}\:{n}_{\mathrm{1}} \:=\:\mathrm{3}{i}\:+\:\mathrm{4}{j}\:+\:{tk} \\ $$$$\mathrm{normal}\:\mathrm{for}\:\mathrm{plane}\:\mathrm{2}\:{n}_{\mathrm{2}} =\:{ki}\:+\:\mathrm{6}{j}\:+\:\mathrm{5}{k} \\ $$$$\mathrm{for}\:\mathrm{parrallel}\:\mathrm{vectors}\:\overset{\rightarrow} {{a}}\:\mathrm{and}\:\overset{\rightarrow} {{b}}\:,\:\overset{\rightarrow} {{a}}=\:{h}\:\overset{\rightarrow} {{b}}\:\mathrm{where}\:{h}\:\mathrm{is}\:\mathrm{a}\:\mathrm{constant}\:{h}\:\in\mathbb{R} \\ $$$$\Rightarrow\:\left(\mathrm{3}{i}\:+\:\mathrm{4}{j}\:+\:{tk}\right)\:=\:{h}\left({ki}\:+\:\mathrm{6}{j}\:+\:\mathrm{5}{k}\right) \\ $$$$\:\:\Rightarrow\:\:\mathrm{4}\:=\:\mathrm{6}{h}\:\Leftrightarrow\:{h}\:=\:\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\mathrm{also}\:\mathrm{3}\:=\:{hk}\:\Rightarrow\:\:{k}\:=\:\frac{\mathrm{3}}{{h}}\:=\:\mathrm{3}\:×\:\frac{\mathrm{3}}{\mathrm{2}}\:\:,\:{k}\:=\:\frac{\mathrm{9}}{\mathrm{2}} \\ $$$$\mathrm{and}\:{t}\:=\:\mathrm{5}{h}\:\Rightarrow\:\:{t}\:=\:\frac{\mathrm{10}}{\mathrm{3}} \\ $$