Calculus and Vectors – How to get an A+
8.5 Cartesian Equation of a Plane
A Normal Equation of a Plane
A plane may be determined by a point
P0 ( x0 , y0 , z 0 ) and a vector perpendicular to the
r
plane n called the normal vector.
B Cartesian Equation of a Plane
Let write the normal vector of a plane in the form:
r
n = ( A, B, C )
Then, the normal equation (1) may be written as:
( x − x0 , y − y0 , z − z 0 ) ⋅ ( A, B, C ) = 0
Ax + By + Cz − Ax0 − By0 − Cz 0 = 0
or:
Ax + By + Cz + D = 0
If P( x, y, z ) is a generic point on the plane, then:
r
P0 P ⊥ n and:
r
P0 P ⋅ n = 0 (1)
This is the normal equation of a plane.
Ex 1. Consider the plane π defined the Cartesian
equation π : 2 x − 3 y + 6 z + 12 = 0 .
a) Find a normal vector to this plane.
r
n = (2,−3,6)
b) Find two points on this plane.
If x = 0 and y = 0 then z = −2 . So (0,0,−2) ∈ π .
If x = 0 and z = 0 then y = 4 . So (0,4,0) ∈ π .
c) Find if the point P(1,2,3) is a point on this plane.
2(1) − 3(2) + 6(3) + 12 = 26 ≠ 0 ⇒∴ P ∉ π
(2)
equation which is called the Cartesian equation of a
plane.
Note. A normal vector to the plane is:
r r r
n = u ×v
r
r
where u and v are the direction vectors of the plane.
Ex 2. Find the Cartesian equation of a plane π that
passes through the points A(1,−1,0) , B(0,0,1) , and
C (0,−2,1) .
r
r
u = AB = (−1,1,1); v = AC = (−1,−1,1)
r r
r
r
r
i
j
k
i
j
r r r
− 1 1 = (2,0,2) = ( A, B, C )
n = u × v = −1 1 1
−1 −1 1
−1 −1
2x + 2z + D = 0
A(1,−1,0) ∈ π ⇒ 2(1) + 2(0) + D = 0 ⇒ D = −2
∴ 2 x + 2 z − 2 = 0 or x + z − 1 = 0
Ex 3. Find parametric and vector equations for the
plane:
π : x − 2 y + 3z − 6 = 0
Ex 4. Find the intersections with the coordinate axes for
the plane π : 3x + 2 y + z − 6 = 0 . Represent the plane
graphically.
Method #1
⎧ x = 6 + 2s − 3t
⎪
; s, t ∈ R
⎨y = s
⎪z = t
⎩
r
r = (6,0,0) + s (2,1,0) + t (−3,0,1), s, t ∈ R
Let A = π ∩ x − axis . y A = z A = 0 ⇒ x A = 2 .
∴ x − int = A(2,0,0) .
Let B = π ∩ y − axis . x B = z B = 0 ⇒ y B = 3 .
∴ y − int = B(0,3,0) .
Let C = π ∩ z − axis . xC = yC = 0 ⇒ zC = 6 .
∴ z − int = C (0,0,6) .
Method #2
A(0,0,2) ∈ π ; B(0,−3,0) ∈ π ; C (6,0,0) ∈ π
r
r
u = AB = (0,−3,−2); v = AC = (6,0,−2)
r
∴ r = (0,0,2) + s (0,−3,−2) + t (6,0,−2); s, t ∈ R
⎧ x = 6t
⎪
⎨ y = −3s
⎪ z = 2 − 2 s − 2t
⎩
8.5 Cartesian Equation of a Plane
©2010 Iulia & Teodoru Gugoiu - Page 1 of 2
Calculus and Vectors – How to get an A+
Ex 5. Find the Cartesian equation of a plane with
x − int = −1 , y − int = 2 , and z − int = −3 .
Ax + By + Cz + D = 0
D
= −1 ⇒ A = D
A
D
D
y − int = − = 2 ⇒ B = −
B
2
D
D
z − int = − = −3 ⇒ C =
C
3
D
D
Dx − y + z + D = 0
2
3
y z
∴ x − + + 1 = 0 or ∴ 6 x − 3 y + 2 z + 6 = 0
2 3
x − int = −
F Angle between two Planes
The angle between two plane s is defined as the
angle between their normal vectors:
r r
n1 ⋅ n 2
cos θ = r r
| n1 || n 2 |
Note. Using this formula, you may get an acute or
an obtuse angle depending on the normal vectors
which are used.
Ex 6. Find the angle between each pair of planes.
a) π 1 : x + 2 y + 3z + 1 = 0 , π 2 : 3 x + 2 y + z + 2 = 0
r
r
n1 = (1,2,3); n2 = (3,2,1)
r r
n ⋅n
(1)(3) + (2)(2) + (3)(1)
10
cosθ = r 1 r2 =
=
2
2
2
2
2
2
| n1 || n2 |
14
1 + 2 + 3 3 + 2 +1
∴θ = cos −1 (10 / 14) ≅ 44.42°
b) π 1 : x + y + z + 1 = 0 , π 2 : x − y − 1 = 0
r
r
n1 = (1,1,1); n2 = (1,−1,0)
r r
n ⋅n
(1)(1) + (1)(−1) + (1)(0)
0
=
=0
cosθ = r 1 r2 =
2
2
2
2
2
2
| n1 || n2 |
3 2
1 + 1 + 1 1 + (−1) + 0
∴θ = 90°
Reading: Nelson Textbook, Pages 461-468
Homework: Nelson Textbook: Page 468 #1, 5, 7, 8, 9a, 11, 13, 17
8.5 Cartesian Equation of a Plane
©2010 Iulia & Teodoru Gugoiu - Page 2 of 2