Equation of planes can expressed in either vector or cartesian form. Vector
equation is the one containing only vectors. Cartesian equation is in the form:
A x + B y + C z = D Ax+By+Cz=D A x + B y + C z = D .
Contains a point and parallel to 2 vectors
Suppose a plane:
is parallel to both a ‾ \underline{a} a and b ‾ \underline{b} b where
a × b ≠ 0 a\times b \neq 0 a × b = 0
contains r 0 ‾ = x 0 i ‾ + y 0 j ‾ + z 0 k ‾ \underline{r_0}=x_0\underline{i}+y_0\underline{j}+z_0\underline{k} r 0 = x 0 i + y 0 j + z 0 k
Equation for the plane is:
r ‾ = r 0 ‾ + s a ‾ + t b ‾ ; s , t ∈ R \underline{r}=\underline{r_0}+s\underline{a}+t\underline{b}\;;\;s,t\in\mathbb{R}
r = r 0 + s a + t b ; s , t ∈ R
Contains a point and normal is given
Suppose a plane:
contains r 0 ‾ = x 0 i ‾ + y 0 j ‾ + z 0 k ‾ \underline{r_0}=x_0\underline{i}+y_0\underline{j}+z_0\underline{k} r 0 = x 0 i + y 0 j + z 0 k
has a normal n ‾ \underline{n} n
Equation for the plane is:
( r ‾ − r 0 ‾ ) ⋅ n ‾ = 0 (\underline{r}-\underline{r_0})\cdot\underline{n} = 0
( r − r 0 ) ⋅ n = 0
Contains 3 points
Suppose a plane contains r 0 , r 1 , r 2 r_0,r_1,r_2 r 0 , r 1 , r 2
(r 0 ‾ , r 1 ‾ , r 2 ‾ \underline{r_0},\underline{r_1},\underline{r_2} r 0 , r 1 , r 2 are the position vectors of
respectively).
( r ‾ − r 1 ‾ ) ⋅ [ ( r 1 ‾ − r 0 ‾ ) × ( r 1 ‾ − r 2 ‾ ) ] = 0 (\underline{r}-\underline{r_1})
\,
\cdot
\,
\Big[
(\underline{r_1}-\underline{r_0})
\times
(\underline{r_1}-\underline{r_2})
\Big] = 0
( r − r 1 ) ⋅ [ ( r 1 − r 0 ) × ( r 1 − r 2 ) ] = 0
Normal to a plane
Suppose a x + b y + c z = d ax+by+cz=d a x + b y + cz = d is a plane.
n ‾ = a i ‾ + b j ‾ + c k ‾ \underline{n}=a\underline{i}+b\underline{j}+c\underline{k} n = a i + b j + c k is a normal to the
plane.
Angle between 2 planes
Consider the two planes:
A : a 1 x + a 2 y + a 3 z = d A: a_1x+a_2y+a_3z=d A : a 1 x + a 2 y + a 3 z = d
B : b 1 x + b 2 y + b 3 z = d ′ B: b_1x+b_2y+b_3z=d' B : b 1 x + b 2 y + b 3 z = d ′
The angle between the planes ϕ \phi ϕ is given by:
cos ( ϕ ) = n A ‾ ⋅ n B ‾ ∣ n A ∣ ⋅ ∣ n B ∣ = a 1 b 1 + a 2 b 2 + a 3 b 3 ( a 1 2 + a 2 2 + a 3 2 ) ( b 1 2 + b 2 2 + b 3 2 ) \cos(\phi)
=\frac{\underline{n_A}\cdot\underline{n_B}}{|n_A|\cdot|n_B|}
=\frac{a_1b_1+a_2b_2+a_3b_3}{\sqrt{(a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2)}}
cos ( ϕ ) = ∣ n A ∣ ⋅ ∣ n B ∣ n A ⋅ n B = ( a 1 2 + a 2 2 + a 3 2 ) ( b 1 2 + b 2 2 + b 3 2 ) a 1 b 1 + a 2 b 2 + a 3 b 3
Here n A ‾ , n B ‾ \underline{n_A},\underline{n_B} n A , n B are normal to the planes A , B A,B A , B .
Shortest distance from a point
Consider the plane a x + b y + c z = d ax+by+cz=d a x + b y + cz = d .
distance = ∣ ( r 1 ‾ − r 0 ‾ ) ⋅ n ‾ ∣ ∣ n ‾ ∣ \text{distance}=
\frac{
\big\lvert
(\underline{r_1}-\underline{r_0})\cdot\underline{n}
\big\rvert
}{
\lvert{\underline{n}}\rvert
}
distance = ∣ n ∣ ( r 1 − r 0 ) ⋅ n
n ‾ \underline{n} n is a normal to the plane
r 0 ‾ \underline{r_0} r 0 is the position vector of any known point on the plane
r 1 ‾ \overline{r_1} r 1 is the position vector to the arbitrary point
Intersection
In 3D, to prove 2 planes intersect, it has to be proven that there is a point
satisfiying both of the planes.
Of 2 planes
Can either be a:
Plane - when the planes coincicde
Line - otherwise
Equation of the line of intersection can be found by:
Solving y , z y,z y , z with respect to x x x
Subject x x x and symmetric form can be found
Of 3 planes
Can either be a:
Plane - when the planes coincide
Line - when the lines of intersection between the planes pairwise coincide
Point - otherwise
First pairwise intersection of the planes must be found. And then intersection
of those 2 can be found.