This can be rewritten this way:[br][math]\vec{c}\left(t\right)=\left(h,k\right)+\left(R\cos t\right)\vec{i}+\left(R\sin t\right)\vec{j}[/math][br][br]In [math]\mathbb{R}^3[/math] we can shift our circle out of the [math]xy-[/math]plane by finding two non-zero orthogonal vectors [i]in[/i] the plane containing our circle. If [math]\left(h,k,j\right)[/math] is the center of the circle and [math]\vec{v}[/math] and [math]\vec{w}[/math] are two orthogonal vectors in the plane containing our circle, then we have the following parameterization:[br][br][math]\vec{c}\left(t\right)=\left(h,k,j\right)+\left(R\cos t\right)\frac{\vec{v}}{\left\left||\vec{v}\right|\right|}+\left(R\sin t\right)\frac{w}{\left|\left|\vec{w}\right|\right|}[/math][br][br]The GeoGebra file below illustrates this approach. Have a conversation with a partner about why this works. Why are we dividing by [math]\left|\left|\vec{v}\right|\right|[/math] and [math]\left|\left|\vec{w}\right|\right|[/math]? What happens if [math]\vec{v}[/math] and [math]\vec{w}[/math] are not orthogonal?