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This was an extra credit question on my Calculus final.

Parametrize the circle lying in the plane with normal vector $(1, 1, -2)$ with center at $$\Big(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\Big)$$ and passing through the origin.

I figured out that looking down the $z$ axis the "circle" would resemble a ellipse, while along the $x$ or $y$ axis it would look like a line with the values bobbing back and forth periodically but I wasn't able to convince myself of a good answer. I came up with $$\Big(\sin(t)\cos^{-1}\big(\tfrac{1}{\sqrt{3}}\big), \sin(t)\sin^{-1}\big(\tfrac{1}{\sqrt{3}}\big), (\text{some linear function})\Big)$$

Surb
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SwimMaster
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4 Answers4

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First $r=1$.

Next we get $(\frac1{\sqrt3},\frac1{\sqrt3},\frac1{\sqrt3})+\cos t\cdot\frac1{\sqrt5} (2,0,1)+\sin t\cdot \frac1{\sqrt{30}}(-1,5,2)$ by looking at this answer.

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Any circle can be parameterized as $C + (A-C) \cos t + (B-C)\sin t$, where $C$ is the circle’s center and the segments $\overline{CA}$ and $\overline{CB}$ are perpendicular radii. This can be derived by applying a similarity transformation to the unit circle $(\cos t,\sin t,0)$.

You’ve been given $C = \frac1{\sqrt3}(1,1,1)$ and that the origin lies on the circle, so you have one of the radii, either $C$ or $-C$, as you prefer. You’ve also been given a normal to the plane that the circle lies on, so you can find a perpendicular radius via a cross product: $${(1,1,-2) \over \|(1,1,-2\|} \times \left(\frac1{\sqrt3},\frac1{\sqrt3},\frac1{\sqrt3}\right) = \left(\frac1{\sqrt2},-\frac1{\sqrt2},0\right).$$ Putting this all together, one possible parameterization is $$\left(\frac1{\sqrt3},\frac1{\sqrt3},\frac1{\sqrt3}\right) + \cos t \left(\frac1{\sqrt3},\frac1{\sqrt3},\frac1{\sqrt3}\right) + \sin t \left(\frac1{\sqrt2},-\frac1{\sqrt2},0\right) = \left({1+\cos t\over\sqrt3}+{\sin t\over\sqrt2}, {1+\cos t\over\sqrt3}-{\sin t\over\sqrt2}, {1+\cos t\over\sqrt3}\right).$$

This same method can be used to parameterize an ellipse, too, although instead of perpendicular radii you need a pair of conjugate radii.

amd
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You have the center and a point, thus you have the radius, which is $1$. And you have the vector $\vec {CO}=\bf r =(-1/\sqrt{3},-1/\sqrt{3},-1/\sqrt{3})$.

Take the cross product of the normal vector with this $\bf t =\bf n \times \bf r$: you have two orthogonal vectors in the plane of the circle.

Then $\vec{OP}= \vec{OC}+ \bf r \cos \theta+ \bf t \sin \theta=\bf r (\cos \theta-1)+ \bf t \sin \theta$

G Cab
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If your circle was in the $xy-$plane with center at $\mathbf{p}$, then one can paramterize it by $\mathbf{r}(t)=\mathbf{p}+(a\cos t) \hat{\mathbf{i}}+ (a \sin t)\hat{\mathbf{j}}$ (with $0 \leq t \leq 2\pi$), where $a$ is the radius of the circle.

But your circle is in a different plane. So we first need to find two unit, mutually orthogonal vectors $\bf{u}$ and $\bf{v}$ in the plane of the given circle (so that they can serve as the replacements of $\hat{\mathbf{i}}$ and $\hat{\mathbf{j}}$).

The equation of the plane is: $x+y-2z=0$. Then we can choose
$$\bf{u}=\rm{\frac{1}{\sqrt{3}}(1,1,1)} \qquad \bf{v}=\rm{\frac{1}{\sqrt{2}}(-1,1,0)}.$$

Observe that the circle passes through the origin, so the radius is $\sqrt{\frac{1}{3}+\frac{1}{3}+\frac{1}{3}}=1$. So the parametric equation of your circle is: $$\mathbf{r}(t)=\mathbf{p}+(1\cos t) \mathbf{u}+ (1 \sin t)\mathbf{v}.$$ Which translates to $$\mathbf{r}(t)=\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)+(\cos t) \frac{1}{\sqrt{3}}(1,1,1)+(\sin t)\frac{1}{\sqrt{2}}(-1,1,0).$$ Simplify it to get $$\mathbf{r}(t)=\left(\frac{1+\cos t}{\sqrt{3}}-\frac{\sin t}{\sqrt{2}}, \frac{1+\cos t}{\sqrt{3}}+\frac{\sin t}{\sqrt{2}}, \frac{1+\cos t}{\sqrt{3}}\right) \quad (\text{with }0 \leq t \leq 2\pi).$$

NOTE: If you choose a bit "nice" vectors, then the answer will be prettier. But I just went with what numbers came to me in the first go.

Anurag A
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