Question Number 68836 by Joel122 last updated on 16/Sep/19
$$\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{graph}\:\mathrm{of} \\ $$$$\boldsymbol{{r}}\:=\:\left(\mathrm{sin}\:{t}\right)\boldsymbol{{i}}\:+\:\left(\mathrm{2cos}\:{t}\right)\boldsymbol{{j}}\:+\:\left(\sqrt{\mathrm{3}}\mathrm{sin}\:{t}\right)\boldsymbol{{k}} \\ $$$$\mathrm{is}\:\mathrm{a}\:\mathrm{circle} \\ $$
Answered by Tanmay chaudhury last updated on 16/Sep/19
$${ix}+{jy}+{kz}=\left({sint}\right){i}+\left(\mathrm{2}{cost}\right){j}+\left(\sqrt{\mathrm{3}}\:{sint}\right){k} \\ $$$${x}={sint} \\ $$$${y}=\mathrm{2}{cost} \\ $$$${z}=\sqrt{\mathrm{3}}\:{sint} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} ={sin}^{\mathrm{2}} {t}+\mathrm{4}{cos}^{\mathrm{2}} {t}+\mathrm{3}{sin}^{\mathrm{2}} {t} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\mathrm{4}\rightarrow{eqn}\:{of}\:{sphere} \\ $$