Parameterizing an oblique circle in 3D

(10 points)

Let

• F be the position in the alphabet of your first initial,
• M be the position in the alphabet of your middle initial, and
• L be the position in the alphabet of your last initial.

Parameterize the circle of radius 2 centered at the origin and contained in the plane

Then, share a Desmos3D link illustrating your result using the techniques I used here:

Be sure to typeset your response! It might look something like this:

You can cntrl-click or alt-click on that expression above to pull up the LaTeX code to generate it and you can use the proofreader to help out as well.

Name Circle Parameter | Desmos

My letters are REG (18,5,7)

My Equation: \vec{p}(t) = \frac{2\cos(t)}{\sqrt{349}}\langle 5,18,0\rangle + \frac{2\sin(t)}{\sqrt{125342}}\langle -126,35,329\rangle

My initials where DJL, and my parameterizing equation ended up being:

Here's my Desmos 3D link!: link

The equation of my plane is

5x - 13y + 19z = 0

The equation of a circle of radius 2 centered at the origin in this plane is

Here is a desmos link to my graph

Here's the link for my Parameterized Circle 3D Graph

My intitials were JGB and my equation was

My equation was

The equation of my plane would be

From this I was able to get my normal vector

To find my \mathbf{u} vector I made

based on

x=0,
y=13,
z=11.
Now that I have my normal vector and my vector u I took the cross product to find my vector v:

Now I have vector u and vector v, I found the magnitude of both of them in order to normalize them:
|\mathbf{u}| = \sqrt{290},
|\mathbf{v}| = \sqrt{103725}.
So my final equation is

My initials are NPC (14, 16, 3).

So my equation came out to be:

@User 045
Do you have a link to what this looks like in Desmos 3D?

Added to original post!

my intials are JKG, so 10,11,7
the plane is:
10x-11y+7z=0
the equation becomes

heres what it looks like in desmos!

My initials are KMM and my equation is:

Here is the link to my desmos Calc 3 | Desmos

My initials are JRH, (10,18,8), so my equation for the plane looks like:

I took \langle 10,-18,8\rangle as my normal vector \vec{n} from the plane. Then, I found \vec{u} as \langle 1,1,1\rangle since \vec{n}\cdot\vec{u}=0. To find \vec{v}, I computed \vec{u}\times\vec{n}=\langle -26,-2,28\rangle. Then, to set the vectors as unit vectors, I divided them by their magnitudes. Since we're looking to parameterize a circle of radius = 2, multiply \vec{u} and \vec{v} by 2cos(t) and 2sin(t) respectively.

Thus, the parameterization of the circle I found is:

My initials are MCS. In relation to the alphabet, the numbers are (13,3,19).

From this I got my normal vector which is essentially the same as the plane.

n = \langle 13,-3,19\rangle.

To get vector u I made

u = \langle 3,13,0\rangle.

Then I took the cross product of both vectors n and u to find vector v:

n \times u = \langle -247,57,178\rangle.

Then I found the magnitudes of both u and v.

\lVert\textbf{u}\rVert = \sqrt{178}
\lVert\textbf{v}\rVert = \sqrt{95942}

Therefore,

My equation is:

The link to my graph is here:

My initials are OAF and my equation is

Desmos graph

Visit my graph at: Bennett Name Circle Desmos Graph

My name letters (BAM) are determined by the normal vector

My Equations:

Circle determined by function \mathbf{c}(t)=2\cos\left(t\right)\mathbf{u}+2\sin\left(t\right)\mathbf{v}.

My initials are whn.

Vector \mathbf{n} = \langle 8,\;-23,\;14\rangle.
Vector \mathbf{u} = \langle 23,\;8,\;0\rangle.
Vector \mathbf{v} = \langle -112,\;322,\;593\rangle.

My equations:

Circle: c(t) = l(0) + 2\cos(t)\mathbf{u} + 2\sin(t)\mathbf{v}.

My initials are RSD (18,19,4)

My equation:p(t) = ((108/41)+57\cos t+12\sin t-(114/41)+54\cos t-54\sin t)

Mylink

My initials are FSV: (6,19,22)
My equation is:

Desmos3D link: Circular Motion 3D Graph

My initials are AJD (1,10,4).
My equation is

This is the link to my model: Parameterizing an obique circle | Desmos

Here is the link to the Desmos graph: Circle Name Graph | Desmos

My initials are: CMJ (3, 13, 10)

And the parameterizing equation is:

My initials are EVV and my equation is:

Here is the link to my Desmos: Parameterizing circle in 3D | Desmos

My letters were ABA (1,2,1).

My equation was vector \mathbf{p}(t) = (-5,2,1)\frac {2\sin(t)}{\sqrt{30}} + (0,1,-2)\frac {2\cos(t)}{\sqrt{5}}.
This is my Desmos link: Circular motion in 3D | Desmos.

Letters: BMD

My initials are SDT and my equation is:
\vec{p}(t) = \frac{2\cos(t)}{\sqrt{416}} \langle 0,4,20 \rangle + \frac{2\sin(t)}{\sqrt{323232}} \langle -416, -76, 380 \rangle
Here's the link to my Desmos graph:

My initials are: SAV
So, my equation of my plane is

My normal vector is

Then I found

Then I took the cross product of the two vectors and got

Then, to make them unit vectors I found the magnitudes of vectors u and v. Therefore, my parameterization of my circle is:

Here is the link to my graph:

My initials are ZSO.
This is my equation:

My letters are (A, A, Z) \to \langle 1, 1, 26\rangle.

My equation is:

My letters were NIL (14,-9,12).

My equation ended up being

This is my desmos link

My initials are JWC (10, 23, 3), giving me an initial equation of (10x - 23y + 3z = 0). This gave me a vector of (10, -23, 3).
After some looking through the textbook and notes, I got a u vector of (0, 3, 23), with a magnitude of \sqrt{538}. The cross product from my initial vector and the u vector was (-538, -230, 30), with a magnitude of \sqrt{343,244}. Finally, I derived my final equation:

\vec{𝑝}(𝑡)=\frac{2\cos\left(t\right)}{\sqrt{538}}\left(0,3,23\right)+\frac{2\sin\left(t\right)}{\sqrt{343244}}\left(-538,-230,30\right)

Here is the link to my Desmos page!

My Initials are DMH, so I used the equation

The parametrization of my circle is

and this is the link to my graph in Desmos: Desmos Link.

My initials are ALR
c(t)=

My initials are AES: (1, 5, 19)

My equation is:

Desmos link:

My initials are JDW (10, 4, 23), which gives me an initial equation of:

The parameterization of my circle is:

Here's the link to the graph in Desmos!

My initials are ASW: (1,19,23)

My equation is:

My Name: CAS (3,1,19)

My Equation:

@User 047 Can you include a link to a 3D graph of your circle in the plane?

Initials: AJH (1,-10,8)

My initials are TCS, which corresponds to values (20, 3, 19).

The equation for my circle looks like this:

\vec p(t)=\frac{2\cos (t)}{\sqrt{171}}\langle 1,13,1\rangle+\frac{2\sin (t)}{\sqrt{131670}}\langle 250,1,-263\rangle

Here is the link to my Desmos graph: Circular Motion in 3D - Tyler Smith