Unit vector Normal to the plane.

Question

I am new to electromagnetics. I came across a problem where I had to find a find a unit vector normal to a plane. When points are given, it is the cross product, I understood that much. But, When the plane is given as 3x+4y+5z = 12, how do I proceed to find a normal vector for this plane?

My question is "HOW DO I FIND A UNIT VECTOR NORMAL TO THE PLANE 3X+4Y+5Z=12"

Answer 1

If a (hyper-)plane $P$ in $\Bbb R^n$ is given by the equation

$\vec a \cdot \vec x + d = \displaystyle \sum_1^n a_i x_i + d = 0, \tag 1$

where

$0 \ne \vec a = (a_1, a_2, \ldots, a_n)^T \in \Bbb R^n \tag 2$

is a constant vector, and

$\vec x = (x_1, x_2, \ldots, x_n)^T \in \Bbb R^n \tag 3$

represents the variable point in $\Bbb R^n$, then for any two points

$\vec y, \vec z \in P \tag 4$

we have

$(\vec y - \vec z) \cdot \vec a = 0, \tag 5$

since

$\vec a \cdot \vec y + d = 0 = \vec a \cdot \vec z + d, \tag 6$

whence

$\vec a \cdot (\vec y - \vec z) = (\vec a \cdot \vec y + d) - (\vec a \cdot \vec z + d) = 0; \tag 7$

since any vector $\vec w$ which is tangent to $P$ may be expressed in the form

$\vec w = \vec y - \vec z, \; \vec y, \vec z \in P, \tag 8$

(7) shows

$\vec a \cdot \vec w = 0 \Longrightarrow \vec a \bot P; \tag 9$

since we assume

$\vec a \ne 0, \tag{10}$

we may form a unit vector

$\vec n = \dfrac{\vec a}{\Vert \vec a \Vert}, \tag{11}$

whence

$\vec w \cdot \vec n = \dfrac{\vec w \cdot \vec a}{\Vert \vec a \Vert} = 0; \tag{12}$

this shows $\vec n$ is a unit vector with

$\vec n \bot P; \tag{13}$

note that, with $\vec x \in P$,

$\Vert a \Vert \left (\vec n \cdot \vec x + \dfrac{d}{\Vert a \Vert} \right) = \Vert a \Vert \vec n \cdot \vec x + \Vert a \Vert \dfrac{d}{\Vert a \Vert} =\vec a \cdot \vec x + d = 0, \tag{14}$

so that we may in fact re-write (1) in the form

$\vec n \cdot \vec x + \dfrac{d}{\Vert a \Vert} = 0; \tag{15}$

in fact, if we write this as

$\vec n \cdot \vec x = -\dfrac{d}{\Vert a \Vert}, \tag{16}$

we see that it may be interpreted as saying that the projection of any $\vec x \in P$ along $\vec n$ has the same magnitude $\vert d \vert / \Vert a \Vert$, which is the distance from the origin to $P$.

In the present case, we have

$\vec a = (3, 4, 5)^T, \tag{17}$

$\Vert \vec a \Vert = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{50}, \tag{18}$

and we may take

$\vec n = \left ( \dfrac{3}{\sqrt{50}}, \dfrac{4}{\sqrt {50}}, \dfrac{5}{\sqrt{50}} \right )^T. \tag{19}$

Answer 2

The vector normal to the plane is given by $$\vec{n}=[3;4;5]$$ and the unit Vector is $$\vec{n}^0=\frac{[3;4;5]}{\sqrt{9+16+25}}$$

Answer 3

Let consider at first a plane through the origin that is

$$ax+by+cz=0$$

and the two vectors

• $OP=(x,y,z)$ indicating a point in the plane
• $n=(a,b,c)$

then the plane equation can be interpreted as the dot product

$$OP\cdot n=ax+by+cz=0$$

which indicates that $n$ is perpendicular to the plane.

Since a plane with general equation

$$ax+by+cz=d$$

is parallel to the plane $ax+by+cz=0$ (we can show that by a translation) also in general a normal to the plane is given by $n=(a,b,c)$.