Question

find the equation of plane in normal form

Answer 1

A Normal Vector usually denoted , is a vector that is perpendicular to a plane. The Point-Normal Form Equation of the plane is where is any normal vector to and is any point.

Answer 2

Solution:

The equation of plane in the normal form can be identified by two factors.

1. "Normal to the plane"

2. "Distances from origin to a plane”

When the plane is in normal form then its vector equation will be,

$\vec{r} \cdot \widehat{n} = {d}$

Where,

$\vec{r}$ is position vector

$\widehat{n}$ is the unit normal joining along the origin normal plane and the variable d is perpendicular plain distance

If p(x, y, z) is any point and O is the origin. Then,

$\vec{o p}=\vec{r}=x \hat{\imath}+y \hat{\jmath}+z \hat{k}$

Now, direction cosines at $\hat{n}$ are "l, m and n".

So, we have,

$\hat{n}=l \hat{\imath}+m \hat{\jmath}+n \hat{k}$

From $\vec{r} \hat{n}=d$, we know that,

$(x \hat{\imath}+y \hat{\jmath}+z \hat{k}) \cdot(l \hat{\imath}+m \hat{\jmath}+n \hat{k})=d$

Thus, in an equation of plane its Cartesian form is

$lx + my + nz = d$