calculusparametricplane-curves
The function $r(t)$ traces a circle. Determine the radius, center, and plane containing the circle parametrized as $ r(t)=−6i+ (9\cos(t))j+ (9\sin(t))k.$ Enter the equation of the plane in the format $ax+by+cz+d=0.$
I already found the radius $(r = 9)$ & the center $(-6, 0, 0)$ but I don't know how to find the plane.
take a hard look at picture take some fixed point anywhere on the circle and when the circle is rolled around the curve drawn by the the fixed point is the TROCHOID
in the diagram i have drawn i labelled it as point D some where on circle also i have taken the angle to be "a"
from the diagram you can see that point D is online CP and P started initially from origin and it has travelled by an angle of "a" so PR=OR= ra ( PCR is a sector with angle a and arc length ra)
our job is to find the x and y co-ordinates of D which is making the curve
from triangle CDQ
CD=d which is given in our question
CQ= d $cosa$ and DQ= d $sina$
now x-coordinate of point D= OR- DQ
SO x= r*a-d $sina$
and y- coordinate of point D = CR- QR
y=r- d $cosa$
hence the parametric representation of the curve is
x= r*a-d $sina$
y=r- d $cosa$
It does because $x^2(t) + y^2(t) = 25 = 5^2 \Rightarrow r = 5, C = (0,0)$
Best Answer
In the format $ax+by+cz+d=0$, the normal vector to the plane is $(a,b,c)$,
and $(a,b,c)\cdot(x,y,z)+d=0$ for any point $(x,y,z)$ in the plane.
$(a,b,c)$ will be normal to any vector in the plane,
say the vector from the center to any point on the circle.
Therefore, you could take vectors from the center to two points on the circle,
and take their cross-product to get a vector normal to the plane.
That would give you an answer for $(a,b,c)$,
and then you could plug in values $(x,y,z)$ of a point in the plane (on the circle) to solve for $d$.
Alternatively, you could simply recognize that the $x$-component of $r(t)$ has only one value,
and that defines a plane.