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)

\begin{align} (1, 4, 12) \cdot (x - 2, y - 4, z - 8) = 0 \\ 1(x - 2) + 4(y - 4) + 12(z - 8) = 0 \end{align}

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)

\begin{align} \quad \vec{r'}(t) \times \vec{r''}(t) = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & 2t & 3t^2 \\ 0 & 2 & 6t \end{vmatrix} = (12t^2 - 6t^2) \vec{i} - 6t \vec{j} + 2 \vec{k} = (6t^2, -6t, 2) \end{align}

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

(3)

\begin{align} (24, -12, 2) \cdot (x - 2, y - 4, z - 8) = 0 \\ 24(x - 2) -12(y - 4) + 2(z - 8) = 0 \end{align}

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)

\begin{align} \quad [\vec{r'}(2) \times \vec{r''}(2)] \times \vec{r'}(2) \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 24 & -12 & 2 \\ 1 & 4 & 12 \end{vmatrix} = (-144 - 8)\vec{i} - (288 - 2)\vec{j} + (96 +12)\vec{k} = (-152, -286, 108) \end{align}

Therefore the rectifying plane is given by:

(5)

\begin{align} (-152, -286, 108) \cdot (x - 2, y - 4, z - 8) = 0 \\ -152(x - 2) - 286(y - 4) + 108(z - 8) = 0 \\ -76(x - 2) - 143(y - 4) + 54(z - 8) = 0 \\ \end{align}