# Equation of a Plane in Normal Form and Cartesian Form

## Equation of a Plane in Normal Form and Cartesian Form

Plane: A plane is a surface such that if any two points are taken on it, the line segment joining them lies completely on the surface.

A plane is determined uniquely if.

• The normal to the plane and its distance from the origin is given, i.e., the equation of a plane in normal form.
• It passes through a point and is perpendicular to a given direction.
• It passes through three given non – collinear points.

Equation of a Plane in Normal Form:

Consider a plane whose perpendicular distance from the origin is d (d ≠ 0). If $$\overrightarrow{ON}$$ is the normal form the origin to the plane, and  is the unit normal vector along  $$\overrightarrow{ON}$$, then  $$\overrightarrow{ON}=d\hat{n}$$.

Let P be any point on the plane. Therefore, $$\overrightarrow{NP}$$ is perpendicular to $$\overrightarrow{ON}$$.

$$\because \overrightarrow{NP} . \overrightarrow{ON}=0$$ … (i)

Let $$\vec{r}$$ be the position vector of the point P. Then $$\overrightarrow{NP}=\vec{r}-d\hat{n} \left( as \, \overrightarrow{ON}+\overrightarrow{NP}=\overrightarrow{OP} \right)$$,

Therefore, (i) becomes

$$\left( \vec{r}-d\hat{n} \right).d\hat{n}=0$$,

$$\left( \vec{r}-d\hat{n} \right).\hat{n}=0\ \ \ \ \left( d\ne 0 \right)$$,

$$\left( \vec{r}.\hat{n} \right)-d\hat{n}.\hat{n}=0$$,

$$\vec{r}.\hat{n}=d (\because \ \hat{n}.\hat{n}=1)$$ … (ii)

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

Cartesian Form:  Equation (ii) given the vector equation of a plane, where $$\hat{n}$$ is the unit vector normal to the plane.

Let P (x, y, z) be any point on the plane. Then $$\overrightarrow{OP}=\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$$,

Let l, m and n be the direction cosines of . Then $$\hat{n}=l\hat{i}+m\hat{j}+n\hat{k}$$,

Therefore, (ii) gives $$(x\hat{i}+y\hat{j}+z\hat{k})(l\hat{i}+m\hat{j}+n\hat{k})=d$$,

lx + my + nz = d.

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