“Teach A Level Maths”
Vol. 2: A2 Core Modules
55: The Vector Equation of
a Plane
© Christine Crisp
The Vector Equation of a Plane
Module C4
MEI/OCR
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The Vector Equation of a Plane
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.
A
x
n
There is only one
plane through A that
is perpendicular to
the vector.
The Vector Equation of a Plane
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.
A
x
R
x
n
Suppose R is any point
on the plane ( other
than A ).

Then, n . AR  0
n . (r  a )  0

This is the equation of the plane since it is
satisfied by the position vector of any point on
the plane, including A.
The scalar product can be expanded to give
n.r  n.a  0 
n.r  n.a
The Vector Equation of a 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.
A
x
xC
There are an infinite number of vectors perpendicular

to AC. For example, one lies on the plane.
The Vector Equation of a 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.
A
x
xC
There are an infinite number of vectors perpendicular

to AC. For example, one lies on the plane.
Others lie at angles to the plane.

Only one is also perpendicular to AB.
The Vector Equation of a 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.
A
B
x
x
xC
There are an infinite number of vectors perpendicular

to AC. For example, one lies on the plane.
Others lie at angles to the plane.

Only one is also perpendicular to AB.
This one is perpendicular to the plane.
The Vector Equation of a Plane
e.g.1 Find the equation of the plane through the
1

point A(2, 3, 1) perpendicular to
.
 3
 
Solution:
2

n.r  n.a

1 
 3
 
 2
 x  1 
.  y    3
   
 z   2
2
. 3 
 
  1
1 
 3
 
 2
 x
.  y  2  9  2 
 
z
1 
 3
 
 2
 x
.  y  9
 
z
The Vector Equation of a Plane
1 
 3
 
 2
 x
.  y  9
 
z
Calculating the left-hand scalar product gives the
Cartesian form of the equation.
x  3 y  2z  9
The Vector Equation of a Plane
e.g.2 Show that the vector n is perpendicular to
the plane containing the points A, B and C
where
  2
n 1 
 
  6
Solution:
2
a 3 
 
  1
 2
b    3
 
  2
1
c 1 
 
  1
The plane containing A, B and C also


contains the vectors AB and AC
The Vector Equation of a Plane
  2
2
n   1  a 3 
 
 

6
 
  1
 2
1
b    3 c   1 
 
 

2
 
  1
So,

n . AB

2 2  0 


AB  b  a    3   3     6
     
  2   1   1 
 1   2    1

AC  c  a   1    3     2
     
  1   1  0 
 2   1 
  2  0 

    
 1  .   6  0 , n . AC
 1 . 2  0
   




  6   1
  6  0 
n is perpendicular to 2 vectors in the plane so is
perpendicular to the plane.
The Vector Equation of a Plane
SUMMARY
 The vector equation of a plane is given by
n . (r  a )  0 or
n.r  n.a
where a is the position vector of a fixed point on
the plane
n is a vector perpendicular to the plane and
r is the position vector of any point on the plane.
 The Cartesian form is
n1 x  n2 y  n3 z  d
where n1, n2 and n3 are the components of n and
d  n.a
n is called the normal vector
The Vector Equation of a Plane
Exercise
1. Find a vector equation of the plane through
the point A(4, 3,  2) with normal vector
 1 
n    2
 
 3 
2. Find the Cartesian equation of the plane through
the point A(1, 1, 1) perpendicular to the vector
0
n2
 
  1
The Vector Equation of a Plane
1. Plane through the point A(4, 3,  2) with normal
vector
1
Solution:
 
n    2
 
 3 
n.r  n.a

 1 
 1   4 
  2 . r    2 .  3 
 
   
 3 
 3    2

 1 
  2 . r   8
 
 3 
The Vector Equation of a Plane
2. Find the Cartesian equation of the plane through
the point A(1, 1, 1) perpendicular to the vector
Solution:
0
n2
 
  1
0
 0  1
n . r  n . a   2  . r   2  . 1 
 
   
  1
  1 1
 0   x
 2  .  y  1  2 y  z

   
  1  z 
0
 2  .r  1
 
  1
1
The Vector Equation of a Plane
Exercise
 3
3. Show that n   0  is perpendicular to the plane
 
1 
containing the points A(1, 0, 2), B(2, 3, 1) and C(2, 2, 1 ).
Solution:
2
1
1
2  1 








    






AB  3  0  3
AC  2  0 
     
   

1
2

3
     
  1  2
3  1 
3  1 


    
    
n . AB  0 . 3  0 , n . AC  0 . 2  0
   
   
 1    3
 1    3
 1 
 2 
 
  3
n is perpendicular to 2 vectors in the plane so is
perpendicular to the plane.
The Intersection of a Line and a Plane
If a line is not parallel to a plane, it will intersect it.
e.g. Find the point of intersection of the line L and
the plane p given by:
L:
r  2 i  3 j  k  t ( i  2 j  2k )
p : r .( i  2 j  k )  2
Solution: The point of intersection is the point with
vector r satisfying both equations.
Can you see how to find this r ?
ANS: Substitute for r from L into p. Solve to
find t and substitute back into L.
The Intersection of a Line and a Plane
I’ll use column vectors:
2
1 
 1 
L : r   3   t  2 p : r .   2  2
 
 
 

1
2
 
 
 1 
 2t   1 
Subs. into p :  3  2t  .   2  2

  

1

2
t

  1 
2  t  6  4t  1  2t  2

7  t

Substituting in L: r  2i  3 j  k  7 ( i  2 j  2k )
The coordinates are
(  5 ,  11 ,  15 )
The Intersection of a Line and a Plane
Exercise
1. Find the point of intersection of the line L
and the plane p where
L:
 x  3 
 1 
 y   2   t  4 
   
 
 z    1
  3
p:
 x   2
 y  .  3  33
   
 z  1 
2. How can you tell that the following line L and
plane p don’t intersect?
L:
 x  1 
 1 
 y    2  t   3
   
 
 z  1 
 1 
p:
 x  2 
 y  .   1  4
   
 z    5
The Intersection of a Line and a Plane
1.
L:
 x  3 
 1 
 y   2   t  4 
   
 
z

1
   
  3
p:
 x   2
 y  .  3  33
   
 z  1 
Solution:
Subs. in

p:
 3  t   2
 2  4t  .  3  33

  
  1  3t   1 
6  2t  6  12t  1  3t  33
 11t  22  t  2
The Intersection of a Line and a Plane
t2
 x  3 
 1 
 y   2   2  4 
   
 
 z    1
  3
 x  5 
  y    10 
   
 z   7
Subs. in L :

Coordinates are
( 5 , 10 ,  7 )
The Intersection of a Line and a Plane
2. How can you tell that the following line L and
plane p don’t intersect?
L:
 x  1 
 1 
 y    2  t   3
   
 
z
1
   
 1 
p:
 x  2 
 y  .   1  4
   
 z    5
Solution: If we form the
A
x
scalar product of the normal
vector of the plane . . .
 1 
 2 
and the direction vector
  1
p    3

n
of the line . . .
 


1 

5


the result is zero showing
they are perpendicular.
The line and plane can only intersect if the line lies
on the plane. Since (1, 2, 1) is on the line but not on
the plane, the line and plane do not intersect.
Finding Angles
The Angle between a Line and a Plane
We use the scalar product with
• the direction vector of the line and
• the normal of the plane.
cosq 
A
x
q
n
a
n. p
np
( find the acute angle )
p
BUT the angle we
want is a.
a  90   q 
Tip: It’s easy to confuse the procedure
for finding angles in the different
situations so I always do a sketch.
Finding Angles
The angle between 2 planes
These lines . . .
are perpendicular to
the line of intersection
of the planes.
q
Finding Angles
The angle between 2 planes
n2
q
a
n1
q
The quadrilateral has 2
angles adding to 180
so, q  a  180 
Finding Angles
The angle between 2 planes
n2
q
n1
q
•
n1 . n 2
cosq 
n1 n2
The angle between 2 planes is equal to the angle
between the normal vectors to the planes.
I’ve illustrated the obtuse angle because it’s
easier to see. We usually give the acute angle.
Finding Angles
SUMMARY
 The angle between a line L and a plane
by
a  90  q


where
cosq 
p is given
n. p
np
q is the acute angle between the direction vector
of the line, p , and the normal vector to the
plane, n.
 The angle between 2 planes p1 and p2 is the angle
between their normal vectors and is given by
n1 . n 2
cosq 
n1 n2
Always
do na sketch
when
finding
angles.
where
n1 and
are
the
normal
vectors
to the
2
line or plane(s)
do.180 to
planes. IfAny
necessary
subtractwill
from
find the acute angle.
Finding Angles
Exercise
1. Find the angle, to the nearest degree, between
the line and plane given below
L:
 x   2
  1
 y    0  s  4 
   
 
z
1
   
4
p:
 x    1
 y .  0   3
   
z  1 
2. Find the angle, to the nearest degree, between
the planes given below
p1 :
 x    4
 y .  1   4
   
z  1 
p2 :
 x   2
 y  .  3  3
   
 z   0
Finding Angles
Solutions:
1.
L:
 x   2
  1
 y    0  s  4 
   
 
 z  1
4
p:
q
a
p
n
 cosq 
5
2 33
n. p
cosq 
A
x
 x    1
 y .  0   3
   
z  1 
 q  52 
np
a  90  q


 a  38 
Finding Angles
Solutions:
2.
p1 :
 x
 y
 
z
n2
  4
.  1   4 p2 :
 
1
q
n1
q
 x   2
 y  .  3  3
   
 z   0
n1 . n 2
cosq 
n1 n2
5
cosq 
18 13
q  109

Acute angle is

71
The Vector Equation of a Plane
The Vector Equation of a Plane
The following slides contain repeats of
information on earlier slides, shown without
colour, so that they can be printed and
photocopied.
For most purposes the slides can be printed
as “Handouts” with up to 6 slides per sheet.
The Vector Equation of a Plane
SUMMARY
 The vector equation of a plane is given by
n . (r  a )  0 or
n.r  n.a
where a is the position vector of a fixed point on
the plane
n is a vector perpendicular to the plane and
r is the position vector of any point on the plane.
 The Cartesian form is
n1 x  n2 y  n3 z  d
where n1, n2 and n3 are the components of n and
d  n.a
n is called the normal vector
The Vector Equation of a Plane
e.g.1 Find the equation of the plane through the
1

point A(2, 3, 1) perpendicular to
.
 3
 
 2
Solution:
n.r  n.a 

1 
 3
 
 2
1 
 3
 
 2
 x  1 
.  y    3
   
 z   2
 x
.  y  2  9  2 
 
z
2
. 3 
 
  1
1   x 
 3 .  y   9
   
 2  z 
Calculating the left-hand scalar product gives the
Cartesian form of the equation.
x  3 y  2z  9
The Vector Equation of a Plane
The Angle between a Line and a Plane
We use the scalar product with
•
•
the direction vector of the line and
the normal of the plane.
cosq 
A
x
q
n
a
n. p
np
( find the acute angle )
p
BUT the angle we
want is a.
a  90   q 
Tip: It’s easy to confuse the procedure
for finding angles in the different
situations so I always do a sketch.
The Vector Equation of a Plane
The angle between 2 planes
n2
q
n1
q
•
n1 . n 2
cosq 
n1 n2
The angle between 2 planes is equal to the angle
between the normal vectors to the planes.
I’ve illustrated the obtuse angle because it’s
easier to see. We normally give the acute angle.
The Vector Equation of a Plane
SUMMARY
 The angle between a line L and a plane p is given
by
n. p


a  90  q where cosq 
np
q is the acute angle between the direction vector
of the line, p , and the normal vector to the
plane, n.
 The angle between 2 planes p1 and p2 is the angle
between their normal vectors and is given by
n1 . n 2
cosq 
n1 n2
where n1 and n2 are the
normal vectors to the planes.
If necessary subtract from 180 to find acute angle.
Always do a sketch when finding angles.
Any line or plane(s) will do.