# Determining Equations of Normal, Rectifying, and Osculating Planes

We have recently defined three types of planes known as Normal, Rectifying, and Osculating Planes. Let $\vec{r}(t) = (x(t), y(t), z(t))$ be a vector-valued function and $P_0(x_0, y_0, z_0)$ be a point on a curve $C$ generated by the vector-valued function $\vec{r}(t_0)$.

**The Normal Plane of $P$:**is perpendicular to $\hat{T}(t_0) = \hat{N}(t_0) \times \hat{B}(t_0)$ and passes through $P_0(x_0, y_0, z_0)$.

**The Rectifying Plane of $P$**is perpendicular to $\hat{N}(t_0) = \hat{B}(t_0) \times \hat{T}(t_0)$ and passes through $P_0(x_0, y_0, z_0)$.

**The Osculating Plane of $P$**is perpendicular to $\hat{B}(t_0) = \hat{T}(t_0) \times \hat{N}(t_0)$ and passes through $P_0(x_0, y_0, z_0)$.

*We should note that the normal plane of $P$ is perpendicular to $\vec{r'}(t_0)$.*

*The rectifying plane of $P$ is perpendicular to $[\vec{r'}(t_0) \times \vec{r''}(t_0)] \times \vec{r'}(t_0)$.*

*The osculating plane of $P$ is perpendicular to $\vec{r'}(t_0) \times \vec{r''}(t_0)$.*

Also recall from the Equations of Planes in Three Dimensional Space page that the equation of a plane can be given by a vector $\vec{n} = (a, b, c)$ that is normal to the plane and a point $P_0(x_0, y_0, z_0)$ on the plane as $\vec{n} \cdot \vec{P_0P} = 0$.

Let's now look at some examples of finding normal, rectifying, and osculating planes.

## Example 1

**Let $\vec{r}(t) = (t, t^2, t^3)$ be a vector-valued function. Find the normal, osculating, and rectifying plane corresponding to $t = 2$.**

First we note that $t = 2$ corresponds to the point $(2, 4, 8)$.

To find the normal plane corresponding to $(2, 4, 8)$ ($t = 0$) we need to find a vector perpendicular to this plane. We could use the tangent vector $\vec{r'}(2)$ or the unit tangent vector in this case. We will use the tangent vector since it is easier to compute. We note that $\vec{r'}(t) = (1, 2t, 3t^2)$. Thus at $t = 2$, we have that $\vec{r'}(2) = (1, 4, 12)$ Therefore the tangent plane is given by:

(1)Now to find the osculating plane corresponding to $(2, 4, 8)$ ($t = 0$) we need to find a vector perpendicular to this plane. We will use the vector $\vec{r'}(2) \times \vec{r'}(2)$ which is parallel to $\hat{B}(2)$. We note that $\vec{r''}(t) = (0, 2, 6t)$ and so:

(2)So $\vec{r'}(2) \times \vec{r''}(2) = (24, -12, 2)$. Therefore the osculating plane is given by:

(3)Now to find the rectifying plane corresponding to $(2, 4, 8)$ $t = 2$), we need to find a vector perpendicular t to this plane. We will use $[\vec{r'}(2) \times \vec{r''}(2)] \times \vec{r'}(2)$ which is perpendicular to $\hat{N}(2) = \hat{B}(2) \times \hat{T}(2)$:

(4)Therefore the rectifying plane is given by:

(5)