Section 6.2. The Equation of a Plane in R3
Recall : In R2 , the set of all points whose coordinates satisfy an equation of the
form ax + by = c (e.g. x + 2y = 3) forms a line. A similar equation in R3 might be
something like ax + by + cz = d; e.g x + 2y − z = 3.
Question: If P is the set of all points in R3 whose coordinates (x, y, z) satisfy
x + 2y − z = 3 (e.g P includes (1,1,0), (4,0,1) etc.) what kind of geometric object
is P ?
Answer: Let ~n = [1, 2, −1] (The components of ~n are the coefficients of x, y and
z in the equation). Let A denote the point (1, 1, 0). Then A belongs to P since
1 + 2(1) − 0 = 3 i.e. the coordinates of A satisfy the equation describing P .
Let B = (x1 , y1 , z1 ) be an arbitrary point of R3 . Then B ∈ P if and only if
x1 + 2y1 − z1 = 3.
~ Its components are given by :
Form the vector AB.
~ = [x1 − 1, y1 − 1, z1 − 0] == [x1 − 1, y1 − 1, z1 ]
AB
Then
~ =
~n.AB
=
=
=
[1, 2, −1].[x1 − 1, y1 − 1, z1 ]
1(x1 − 1) + 2(y1 − 1) − 1(z1 )
x1 − 1 + 2y1 − 2 − z1
x1 + 2y1 − z1 − 3
~ if and only if
In particular ~n ⊥ AB
x1 + 2y1 − z1 = 3
i.e. if and only if B belongs to P .
~ where
Conclusion: P consists of all those points B for which ~n ⊥ AB
(i) A = (1, 1, 0) is a (fixed) point of P
(ii) ~n is the vector [1, 2, −1]
1
pppppp
pppppppppp ppp
pp
p pp
ppp p p
pp
p pp n = [1, 2, −1]
ppp p p ~
pp
p pp
ppp p p
p pp
pp p
r
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A
P
What does this look like physically?
-A plane (a copy of R2 inside R3 ). Vectors lying within the plane are orthogonal
to ~n.
Fact 6.2.1: An equation of the form ax + by + cz = d, where a, b, c and d are
constants, describes a plane in R3 .
Definition 6.2.2: Let P be a plane in R3 . A vector ~n is said to be normal to P
~
(or called a normal vector for P ) if ~n is perpendicular to P . This means ~n ⊥ AB
whenever A and B are points in P .
Remark 6.2.3: If the equation of P is ax + by + cz = d, then the vector ~n = [a, b, c]
is normal to P . Furthermore, any vector which is normal to P is a non-zero scalar
multiple of this ~n.
Note: The vector [1, 2, 3] is normal to both the planes P1 : x + 2y + 3z = 1 and
P2 : x + 2y + 3z = 20.
This means that these planes are parallel.
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qpppppppppppp
ppppppppp
pppppppppppppp
pppppppppppppp
pppppppppppppp
pppppppppppppp
p p p p p p p p p p p p p p ppppp
ppppppppppppppppppppppp
2
The normal vector specifies the “orientation” of a plane but not its location in
R3 , in the way that the slope of a line in R2 specifies its orientation but not its
location. To fully specify a plane in R3 , we need both a normal vector and a point
belonging to the plane.
Example 6.2.4∗ : Find the equation of the plane P which contains the point (2, 0, 1)
and to which the vector ~n = [1, 4, −1] is normal.
Solution: Since [1, 4, −1] is normal to P , the equation of P has the form
1x + 4y − 1z = d
for some constant d. The value of d is determined by the fact that the point (2, 0, 1)
belongs to P , hence x = 2, y = 0, z = 1 must satisfy the equation of P . This means
:
1(2) + 4(0) − 1(1) = d =⇒ d = 1
Equation of P : x + 4y − z = 1.
Example 6.2.5∗ : Let P1 be the plane with equation 2x − y + 3z = 2.
(a) Find the equation of the plane P2 which is parallel to P1 and contains the
point (1, −1, −1).
(b) Find the distance between P1 and P2 .
Solution:
(a) The vector ~n = [2, −1, 3] is normal to P1 . Since P2 is parallel to P1 , ~n is normal
to P2 also; P2 has equation
2x − y + 3z = d
Since the point (1, −1, −1) belongs to P2 ,
2(1) − (−1) + 3(−1) = d =⇒ d = 0
Equation of P2 : 2x − y + 3z = 0
(b) Let h be the distance between P1 and P2 .
.
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~n
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P2
~
AB
q
B
3
P1
To find h:1. Choose any point A on P2 e.g. A=(1,-1,-1)
2. Choose any point B on P1
P1 : 2x − y + 3z = 0
Setting y = 0, z = 0 gives x = 1 =⇒ (1, 0, 0) belongs to P1 . Let B =
(1, 0, 0).
~
3. Form the vector AB
~ = [1 − 1, 0 − (−1), 0 − (−1)] = [0, 1, 1]
AB
Then
~
h = ||proj~n AB||
where ~n is a vector normal to both P1 and P2 : ~n = [2, −1, 3].
~
|[0, 1, 1].[2, −1, 3]|
2
~ = |AB.~n| = p
=√
h = ||proj~n AB||
2
2
2
||~n||
14
2 + (−1) + 3
2
Distance between P1 and P2 : √ .
14
Remark: The above method can also be used to find the distance form a point A
~ where ~n is normal to P and Q is
to a plane P in R3 . This is given by ||proj~n QA||
any point on P .
qA
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~
QA
p
Q
q
~n
P
Problem: Suppose A, B and C are points in R3 which are non-collinear (i.e. all
three do not lie on the same line). Then A, B and C are the vertices of a triangle
in R3 . This triangle may be “filled in” to form a “triangular disc”, which can be
extended to form a unique plane P in R3 , in the way that a line segment in R2
can be extended to form a unique line. This plane P is the only plane in R3 which
contains the points A, B and C - just as it takes two distinct points to determine
a line, it takes three (non-collinear) points to determine a plane.
4
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A
C
P
B
If we know the coordinates of A, B and C, how do we find an equation for P ?
Example 6.2.6∗ : (Summer 2001 Q3 (b)) Find the equation of the plane P in R3
which contains the points A(1, 2, 1), B(2, 4, 1) and C(−1, 0, 3).
Partial Solution: We need a normal vector ~n for P .
p p p p p p p p p p p ppppp
ppppppppppppp
p
p
ppp
p
p
p
p
p
p
p
p
ppppp
ppp
......
ppppppppppppp
............
p
p
p
p
ppp
p
p
p
p
.... ~
p
p
p
p
p
... n
p
p
p
p
p
p
ppp
...
ppppppppp ......q
p
...
p
p
p
p
p
ppp
p
p
p
...
p
.
...
pppppppp
............
p
p
p
ppp
p
p
p
...
p
B(2,
4,
1)
p
.
p
p
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p
p
p
...
p
.
p
p
p
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p
ppp
p
.
p
...ppppppp
..
p.....
p
p
p
p
p
..
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ppp
p
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p
p
.
p
p
p
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...
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p
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p
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C(−1, 0, 3) ppppppp
ppp
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ppp
.
pppp ppp
......
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p
p
p
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p
ppp
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ppppppp ppppp
p
p
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p
p
p
ppp
p
p
p
.. .............
p
p
.
p
ppp
.. ......
ppp
..q.......
ppppp ppppppp
ppp
pppppppp pppp
p
p
p
p
p
p
p
p
p
p
p
ppp
pp
ppppp ppppppp
ppp A(1, 2, 1)
ppppppppppppp
p
p
p
p
ppp
p
p
p
p
p
p
ppp p p p p p p p p p p p p p p p p p p p p p
p
~ and AC
~ lie within P :
The vectors AB
~ = [2 − 1, 4 − 2, 1 − 1] = [1, 2, 0]
AB
~ = [−1 − 1, 0 − 2, 3 − 1] = [−2, −2, 2]
AC
5
Thus ~n must be orthogonal to both [1, 2, 0] and [−2, −2, 2]. How do we find a vector
which is orthogonal to both of these?
The answer requires a construction known as the cross product.
6