Equation of A Plane In The Normal Form - Practice Questions & MCQ

Plane

We know that a line is determined by two points. In other words, for any two distinct points, there is exactly one line that passes through those points, whether in two dimensions or three. Similarly, given any three points that do not all lie on the same line, there is a unique plane that passes through these points. Just as a line is determined by two points, a plane is determined by three.

This may be the simplest way to characterize a plane, but we can use other descriptions as well. For example, given two distinct, intersecting lines, there is exactly one plane containing both lines. A plane is also determined by a line and any point that does not lie on the line.

Equation of a plane in normal form

Vector Form

The vector equation of a plane normal to unit vector $\hat{\mathbf{n}}$ and at a distance d from the origin is $\overrightarrow{\mathbf{r}} \cdot \hat{\mathbf{n}}=d$.
Proof:
Let O be the origin and let $(d \neq 0)$ be the length of perpendicular from origin ( O ) to the given plane.
If $\overrightarrow{\mathrm{ON}}$ is the normal from the origin to the plane, and $\hat{\mathbf{n}}$ is the unit normal vector along $\overrightarrow{O N}$. Then $\overrightarrow{O N}=d \hat{\mathbf{n}}$.
Let $P$ be any point on the plane with position vector $\overrightarrow{\mathbf{r}}$, so that $\overrightarrow{\mathrm{OP}}=\overrightarrow{\mathbf{r}}$.

Now, $\overrightarrow{\mathrm{NP}}$ is perpendicular to $\overrightarrow{\mathrm{ON}}$.
$$
\begin{array}{rrl}
\therefore & (\overrightarrow{\mathrm{OP}}-\overrightarrow{\mathrm{ON}}) \cdot \overrightarrow{\mathrm{ON}}=0 & \\
\Rightarrow & (\overrightarrow{\mathrm{ON}}=0 & (\text { as } \overrightarrow{\mathrm{ON}}+\overrightarrow{\mathrm{NP}}=\overrightarrow{\mathrm{OP}}) \\
\Rightarrow & \overrightarrow{\mathbf{r}}) \cdot d \hat{\mathbf{n}}=0 & \\
\Rightarrow & \overrightarrow{\mathbf{r}} \cdot d \hat{\mathbf{n}}-d^2 \hat{\mathbf{n}} \cdot \hat{\mathbf{n}}=0 & \\
\Rightarrow & d \mathbf{r} \cdot \hat{\mathbf{n}}-d^2|\hat{\mathbf{n}}|^2=0 & (\because d \neq 0) \\
\Rightarrow & \overrightarrow{\mathbf{r}} \cdot \hat{\mathbf{n}}-d=0 & \left(\because|\hat{\mathbf{n}}|^2=1\right) \\
\Rightarrow & \overrightarrow{\mathbf{r}} \cdot \hat{\mathbf{n}}=d &
\end{array}
$$

This is the vector form of the equation of the plane.

Cartesian form

$\overrightarrow{\mathbf{r}} \cdot \hat{\mathbf{n}}=d$ is the vector form of the equation of a plane in the normal form where $\hat{\mathbf{n}}$ is the unit vector normal to the plane.
Let $P(x, y, z)$ is any point in the plane.
Then, $\overrightarrow{O P}=\overrightarrow{\mathbf{r}}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}$
Let $\mathrm{I}, \mathrm{m}, \mathrm{n}$ be the direction cosines of $\hat{\mathbf{n}}$. Then
$$
\hat{\mathbf{r}}=l \hat{\mathbf{i}}+m \hat{\mathbf{j}}+z \hat{\mathbf{k}}
$$

Therefore, the equation $\overrightarrow{\mathbf{r}} \cdot \hat{\mathbf{n}}=d$ gives
$$
\begin{array}{r}
(x \hat{i}+y \hat{j}+z \hat{k}) \cdot(\hat{l} \hat{i}+m \hat{j}+n \hat{k})=d \\
\mathbf{l x}+\mathbf{m y}+\mathbf{n z}=\mathbf{d}
\end{array}
$$

This is the Cartesian equation of the plane in the normal form.

NOTE:
If $\vec{r} \cdot(a \hat{i}+b \hat{j}+c \hat{k})=d$ is the vector equation of a plane, then $\mathrm{ax}+\mathrm{by}+\mathrm{cz}=\mathrm{d}$ is the Cartesian equation of the plane, where $\mathrm{a}, \mathrm{b}$ and c are the direction ratios of the normal to the