55: the vector equation of a plane © christine crisp “teach a level maths” vol. 2: a2 core...
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55: The Vector Equation of a Plane
© Christine Crisp
“Teach A Level Maths”
Vol. 2: A2 Core Modules
There are 3 forms of the equation of a plane. We are going to look at 2 of them.
Suppose we have a vector n through a point A.
There is only one plane through A that is perpendicular to the vector.
Ax
n
There are 3 forms of the equation of a plane. We are going to look at 2 of them.
Suppose we have a vector n through a point A.
Ax
n
Rx
Then, 0. ARn
0)(. arn
This is the equation of the plane since it is satisfied by the position vector of any point on the plane, including A.
Suppose R is any point on the plane ( other than A ).
The scalar product can be expanded to give 0.. anrn n . an . r
oThe angle between the vector AR and n = 90��������������
e.g.1 Find the equation of the plane through the point A(2, 3, -1)
perpendicular to
Solution:
231
1 2 1r 3 3 3
2 1 2 . = 2 + 9 - 2 = 9.
nanr . .
9231
.
zyx
1
r 3 92
.
Calculating the left-hand scalar product gives the Cartesian form of the equation.
923 zyx
9231
.
zyx
923 zyx9231
.
zyx
Diagram
It is useful in some problems to know that a vector n will be perpendicular to the plane if it is perpendicular to 2 non-parallel vectors in the plane.
CxA
x
There are an infinite number of vectors perpendicular to AC. For example, one lies on the plane.
It is useful in some problems to know that a vector n will be perpendicular to the plane if it is perpendicular to 2 non-parallel vectors in the plane.
Cx
Others lie at angles to the plane.
Ax
There are an infinite number of vectors perpendicular to AC. For example, one lies on the plane.
Only one is also perpendicular to AB.
It is useful in some problems to know that a vector n will be perpendicular to the plane if it is perpendicular to 2 non-parallel vectors in the plane.
Cx
Others lie at angles to the plane.
B x
There are an infinite number of vectors perpendicular to AC. For example, one lies on the plane.
This one is perpendicular to the plane.Only one is also perpendicular to AB.
Ax
e.g.2 Show that the vector n is perpendicular to the plane containing the points A, B and C where
612
n
132
a
23
2b
111
c
Solution: The plane containing A, B and C also contains the
vectors andAB
AC
abAB
132
23
2
acAC
132
111
021
16
0
612
n
132
a
23
2b
111
c
,0
16
0
612
..
ABn
So,0
021
612
..
ACn
n is perpendicular to 2 vectors in the plane so is perpendicular to the plane.
The vector equation of a plane is given bySUMMARY
where a is the position vector of a fixed point on the plane n is a vector perpendicular to the plane andr is the position vector of any point on the plane.
n is called the normal vector
0)(. arn or r n a n . .
The Cartesian form is
dznynxn 321
where n1, n2 and n3 are the components of n and
and .
Exercise
1. Find a vector equation of the plane through the point with normal vector
32
1n
)1,1,1(A
)2,3,4( A
120
n
2. Find the Cartesian equation of the plane through the point perpendicular to the vector
1.
32
1n
Plane through the point with normal vector)2,3,4( A
Solution:
nanr . . .
32
1
234
32
1 . . r 8
32
1
. r
Diagram
)1,1,1(A2. Find the Cartesian equation of the plane through the point perpendicular to the vector
120
n
Solution:
nanr . .
120
111
120
. . r 11
20
. r
11
20
.
zyx
12 zy
Solution:
3. Show that is perpendicular to the plane
103
n
containing the points A(1, 0, 2), B(2, 3, -1) and C(2, 2, -1 ).
201
122
AC
321
AB
201
132
331
,0
331
103
..
ABn 0
321
103
..
ACn
n is perpendicular to 2 vectors in the plane so is perpendicular to the plane.
Exercise
Perpendicular Distance of a Plane from the Origin
Using the unit normal the plane equation becomes
= |a|1cos = d
d
where d is the perpendicular distance of the plane from the origin
r.n a.n
Perpendicular Distance of a Plane from the Origin
Using the unit normal the plane equation becomes
= |a|1cos = d
d
So if the plane equation is converted into the unit normal plane equation by dividing by the magnitude on n then d represents the perpendicular distance of a plane from the origin
r.n a.n
0
r. 1 11
3
2 2| n | ( 1) 3 10
So the unit normal form is
0
-1
3 11r. =
10 10
11d=
10
Perpendicular distance of the plane from the origin =11
10
Find the distance of the plane below from the origin
Perpendicular Distance Between Two Planes
Find the perpendicular distance of each plane from the origin and hence find the distance between the two planes by subtraction.
Find the perpendicular distance between
0
r. 1 11
3
0
r. 2 3
6
The negative sign means that plane 2 is on the other side of the origin from plane 1.
Plane 1
0
-1
3 11r. =
10 10
11d
10
0
-2
6 -3r. =
40 40
Plane 2-3
d40
The perpendicular distance between the planes =
11 3 11 3
10 40 10 40
0
r. 1 11
3
0
r. 2 3
6
Diagram