Question Number 68836 by Joel122 last updated on 16/Sep/19
![Show that the graph of r = (sin t)i + (2cos t)j + ((√3)sin t)k is a circle](https://www.tinkutara.com/question/Q68836.png)
$$\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{graph}\:\mathrm{of} \\ $$$$\boldsymbol{{r}}\:=\:\left(\mathrm{sin}\:{t}\right)\boldsymbol{{i}}\:+\:\left(\mathrm{2cos}\:{t}\right)\boldsymbol{{j}}\:+\:\left(\sqrt{\mathrm{3}}\mathrm{sin}\:{t}\right)\boldsymbol{{k}} \\ $$$$\mathrm{is}\:\mathrm{a}\:\mathrm{circle} \\ $$
Answered by Tanmay chaudhury last updated on 16/Sep/19
![ix+jy+kz=(sint)i+(2cost)j+((√3) sint)k x=sint y=2cost z=(√3) sint x^2 +y^2 +z^2 =sin^2 t+4cos^2 t+3sin^2 t x^2 +y^2 +z^2 =4→eqn of sphere](https://www.tinkutara.com/question/Q68840.png)
$${ix}+{jy}+{kz}=\left({sint}\right){i}+\left(\mathrm{2}{cost}\right){j}+\left(\sqrt{\mathrm{3}}\:{sint}\right){k} \\ $$$${x}={sint} \\ $$$${y}=\mathrm{2}{cost} \\ $$$${z}=\sqrt{\mathrm{3}}\:{sint} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} ={sin}^{\mathrm{2}} {t}+\mathrm{4}{cos}^{\mathrm{2}} {t}+\mathrm{3}{sin}^{\mathrm{2}} {t} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\mathrm{4}\rightarrow{eqn}\:{of}\:{sphere} \\ $$