# Thread: Show that curve is in a plane

1. ## Show that curve is in a plane

Show that the curve,

$\overline{r(t)} = 4cost\overline{i} + 4sint\overline{j} - sint\overline{k}$

is in a plane, and find the equation for the plane..

If I substitute into the equation of a plane I get,

$a(4cost)+b(4sint)+c(-sint)+d=0$,

I see that it will hold if for example $b=1$ and $c=4$, and so I guess that I have showed that the curve is in a plane, and could use $[0,1,4]$ as my normal vector.

Is there a nice way of doing this so?

Thanks.

2. There are two ways that I know of to show that a curve is in a plane. The basic idea is that you have to show either that all the tangent vectors to the curve are in a plane, or that all the secant vectors are in a plane. If you know calculus, probably the first option is easier; otherwise, do the second option.

1. Show that all the tangent vectors are in a plane. Take the derivative of $\vec{r}(t)$ - call it $\vec{v}(t)$. Show that there exists a nonzero constant vector $\hat{n}$ such that $\vec{v}(t)\cdot\hat{n}=0$ for all $t$.

2. Show that all the secant vectors are in a plane. Produce all vectors of the form $\vec{s}(u,v)=\vec{r}(u)-\vec{r}(v).$ Show that there exists a nonzero constant vector $\hat{n}$ such that $\vec{s}(u,v)\cdot\hat{n}=0$ for all $u,v$.

Unfortunately, your method is not going to work, because your method is trying to show that the radius vectors from the origin to points in the plane are in the plane. That will only work if the origin happens to be one of your radius vectors (i.e., the plane passes through the origin). If the set of radius vectors does not include the origin, then you're going to have to do what I am suggesting.

3. $\vec{v}(t) = -4sint\vec{i} + 4cost\vec{j} - cost\vec{k}$

I let $\vec{n} = [a_1,a_2,a_3]$ and so,

$\vec{v}(t)\cdot \vec{n} = -a_14sint + a_24cost - a_3cost = 0$

This holds for $\vec{n}=[0,1,4]$.

Is this what you meant?

Thanks

4. That looks good to me!