Finding equation of a path in the plane $y=z$

Question

What is the easiest way to see that the path $$\underline{r}:\mathbb{R}\rightarrow \mathbb{R}^3: t \mapsto (\sin\,t,\cos\,t,\cos\,t)$$ traces out an ellipse in the plane $y=z$?

I think first rotating $\mathbb{R}^3$ by $\pi/4$ about the $x$-axis will help but I am not sure how to proceed.

Answer 1

You are almost there. The rotation by $\pi/4$ will yield the path $$(x',y',z')=(\sin t,\sqrt2\cos t,0)$$ in the new coordinates. How do you eliminate $t$? Use $\sin^2t+\cos^2t=1$: $$x'^2+\frac{y'^2}2=1$$

Answer 2

Your path is also contsined in $x^2+y^2=1$, which is a cylinder. So your path is in the intersection of this cylinder and the plane $y=z$. Geometrically, this is an ellipse, since the plane is not paralel to the axis of the cyclinder, so it cuts all generatrices.