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Question Number 222531 by BHOOPENDRA last updated on 29/Jun/25
The foot of the perpendicular from a point of the circle x^2 +y^2 =1,z=0 to the plan 2x+3y+z=6 lie on curve−−−−−−
$${The}\:{foot}\:{of}\:{the}\:{perpendicular}\:{from}\: \\ $$$${a}\:{point}\:{of}\:{the}\:{circle}\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{1},{z}=\mathrm{0} \\ $$$${to}\:{the}\:{plan}\:\mathrm{2}{x}+\mathrm{3}{y}+{z}=\mathrm{6}\:{lie}\:{on}\:{curve}−−−−−− \\ $$
Answered by mr W last updated on 29/Jun/25
P(cos t, sin t, 0) Q(u, v, w) ((u−cos t)/2)=((v−sin t)/3)=((w−0)/1)=s x=2s+cos t y=3s+sin t z=s 2(2s+cos t)+3(3s+sin t)+s=6 ⇒s=((6−2 cos t−3 sin t)/(14)) { ((x=((6+5 cos t−3 sin t)/7))),((y=((18−6 cos t+5 sin t)/(14)))),((z=((6−2 cos t−3 sin t)/(14)))) :}
$${P}\left(\mathrm{cos}\:{t},\:\mathrm{sin}\:{t},\:\mathrm{0}\right) \\ $$$${Q}\left({u},\:{v},\:{w}\right) \\ $$$$\frac{{u}−\mathrm{cos}\:{t}}{\mathrm{2}}=\frac{{v}−\mathrm{sin}\:{t}}{\mathrm{3}}=\frac{{w}−\mathrm{0}}{\mathrm{1}}={s} \\ $$$${x}=\mathrm{2}{s}+\mathrm{cos}\:{t} \\ $$$${y}=\mathrm{3}{s}+\mathrm{sin}\:{t} \\ $$$${z}={s} \\ $$$$\mathrm{2}\left(\mathrm{2}{s}+\mathrm{cos}\:{t}\right)+\mathrm{3}\left(\mathrm{3}{s}+\mathrm{sin}\:{t}\right)+{s}=\mathrm{6} \\ $$$$\Rightarrow{s}=\frac{\mathrm{6}−\mathrm{2}\:\mathrm{cos}\:{t}−\mathrm{3}\:\mathrm{sin}\:{t}}{\mathrm{14}} \\ $$$$\begin{cases}{{x}=\frac{\mathrm{6}+\mathrm{5}\:\mathrm{cos}\:{t}−\mathrm{3}\:\mathrm{sin}\:{t}}{\mathrm{7}}}\\{{y}=\frac{\mathrm{18}−\mathrm{6}\:\mathrm{cos}\:{t}+\mathrm{5}\:\mathrm{sin}\:{t}}{\mathrm{14}}}\\{{z}=\frac{\mathrm{6}−\mathrm{2}\:\mathrm{cos}\:{t}−\mathrm{3}\:\mathrm{sin}\:{t}}{\mathrm{14}}}\end{cases} \\ $$
Answered by BHOOPENDRA last updated on 11/Jul/25
P(x,y,0),O(0,0,0) Q(α,β,γ) (√(x^2 +y^2 )) =1 P(cos θ,sin θ,0) ((α−cos θ)/2)=((β−sin θ)/3)=((γ−0)/1)=(((2cosθ+3sinθ−6))/(14)) 14α=10cos θ−6sinθ+12 14β=−6cosθ+5sin θ+18 70α+84β=14cosθ+168 5α+6β=cos θ+12 cos θ=5α+6β−12 14β=−30α−36β+72+5sin θ+18 6α+10β−18=sinθ (5x+6y−12)^2 +(6x+10y−18)^2 =1 z=6−2x−3y
$${P}\left({x},{y},\mathrm{0}\right),{O}\left(\mathrm{0},\mathrm{0},\mathrm{0}\right)\:\:{Q}\left(\alpha,\beta,\gamma\right) \\ $$$$\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:=\mathrm{1}\:\:\:\:\:\:\: \\ $$$${P}\left({c}\mathrm{os}\:\theta,\mathrm{sin}\:\theta,\mathrm{0}\right) \\ $$$$\frac{\alpha−\mathrm{cos}\:\theta}{\mathrm{2}}=\frac{\beta−\mathrm{sin}\:\theta}{\mathrm{3}}=\frac{\gamma−\mathrm{0}}{\mathrm{1}}=\frac{\left(\mathrm{2}{cos}\theta+\mathrm{3}{sin}\theta−\mathrm{6}\right)}{\mathrm{14}} \\ $$$$\mathrm{14}\alpha=\mathrm{10cos}\:\theta−\mathrm{6sin}\theta+\mathrm{12} \\ $$$$\mathrm{14}\beta=−\mathrm{6cos}\theta+\mathrm{5sin}\:\theta+\mathrm{18} \\ $$$$\mathrm{70}\alpha+\mathrm{84}\beta=\mathrm{14cos}\theta+\mathrm{168} \\ $$$$\mathrm{5}\alpha+\mathrm{6}\beta=\mathrm{cos}\:\theta+\mathrm{12} \\ $$$$\mathrm{cos}\:\theta=\mathrm{5}\alpha+\mathrm{6}\beta−\mathrm{12} \\ $$$$\mathrm{14}\beta=−\mathrm{30}\alpha−\mathrm{36}\beta+\mathrm{72}+\mathrm{5sin}\:\theta+\mathrm{18} \\ $$$$\mathrm{6}\alpha+\mathrm{10}\beta−\mathrm{18}={sin}\theta \\ $$$$\left(\mathrm{5}{x}+\mathrm{6}{y}−\mathrm{12}\right)^{\mathrm{2}} +\left(\mathrm{6}{x}+\mathrm{10}{y}−\mathrm{18}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$${z}=\mathrm{6}−\mathrm{2}{x}−\mathrm{3}{y}\: \\ $$
Commented by BHOOPENDRA last updated on 11/Jul/25

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