Vectors
Section 1: Lines and planes
Notes and Examples
These notes contain subsections on
 Reminder: notation and definitions
 Equation of a line
 The intersection of two lines
 Finding the equation of a plane
 The equation of a plane through three points
 Finding the intersection of a line and a plane
 Finding the distance of a point from a plane
 The angle between a line and a plane
 The angle between two planes
Reminder: notation and definitions
In typescript algebraic letters representing vectors are written in bold. When
writing by hand you cannot easily write in bold, so the convention is that you
underline letters representing vectors. It is very important that you underline
vectors to avoid confusing them with scalar quantities.
e.g. typescript a = hand-written a .
For a vector between two points, say L and M, write the letters representing
the points next to one another with an arrow above.
e.g. vector between point L and point M = LM
It is usual to denote a point by a capital letter and the position vector of that
point by the same letter in lower-case.
e.g. we could write:
l = position vector of point L
m = position vector of point M
So
LM  m  l
The modulus of a vector is the magnitude of the vector, or the length of the
line representing the vector. The modulus of a vector a is written as a and
can be found using Pythagoras’ theorem.
e.g. for the vector a  a1i  a2 j  a3k , a  a12  a22  a32 .
The scalar product of two vectors a and b is written as a.b and defined by
a.b  a b cos
where  is the angle between the vectors a and b.
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The equation of a line
A line is defined by a vector in the direction of the line, and a point on the line.
If a point on the line has position vector a, and the direction vector of the line
is d, then the vector equation of the line is
 a1 
 d1 
 
 
r   a2     d 2 
or
r  a  d
a 
d 
 3
 3
where  is a parameter. Different values of  correspond to different points on
the line.
This vector equation can be written in parametric form:
x  a1   d1 , y  a2   d 2 , z  a3   d3 .
By rearranging each of these parametric equation to give expressions for ,
and equating these three expressions, you obtain the equivalent Cartesian
equation of the line:
x  a1 y  a2 z  a3


.
d1
d2
d3
The intersection of two lines
In two dimensions, two non-parallel lines always intersect. In three
dimensions, however, two lines may not intersect even if they are not parallel.
Lines which do not intersect are called skew lines.
You can find out if two lines intersect simply by attempting to find a point of
intersection, as shown in the example below.
Example 1
 4
5
 5 
2






 
Lines l1 and l2 have vector equations r   3     2  and r   4     1
0
 3
1
0
 
 
 
 
respectively. Do these lines intersect?
Solution
The lines are clearly not parallel as their directions are not parallel (if they were then
5
2
 
 
their direction vectors  2  and  1 would be multiples of one-another).
 3
0
 
 
As they are not parallel, we must establish whether or not they are skew.
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If the lines meet then their parametric equations can be solved simultaneously i.e.
there will exist values of  and  which will satisfy each of these equations
simultaneously:
4  5  5  2 (A)
3  2  4  
(B)
3  1
(C)
Note that different symbols must be
used to denote the parameter in the
two lines. This must be done in order
to avoid confusion. If the same
symbol were used it would imply that
the parameters in each line always
have equal values, which is certainly
not true.
1
1
(C)    ; substitute in (B)    .
3
3
1
1
17 13
.
and   into (A)  
3
3
3
3
This is clearly a contradiction so the equations cannot be solved simultaneously and
the lines must be skew.
Substituting  
In the example above, if the lines did meet, the point of intersection could be
found by substituting one or other of the parameters into the appropriate
parametric equation.
Finding the equation of a plane
The equation of a plane is defined in terms of a vector n normal to the plane.
The vector equation of a plane is
r.n  d
where d is a constant, determined by the position of the plane in space. If you
know the normal vector and the coordinates of a point on the plane, you can
easily find the value of d by substitution.
The equivalent Cartesian equation of the line is
n1 x  n2 y  n3 z  d .
Example 2
(i) Write down the equation of the plane through the point (4, 5, -2) given that the
2
 
vector  1 is perpendicular to the plane.
3
 
(ii) Verify that the point (2, 4, -1) also lies on the plane.
Solution
(i) The equation of the plane is n1 x  n2 y  n3 z  d  0 where d  a  n
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2
 4
 
 
n   1 and a   5 
 2 
3
 
 
 4 2
   
a.n   5    1  4  2  5  (1)  (2)  3  8  5  6  3
 2   3 
All planes with these coefficients of x, y
   
and z will be parallel...
Now d  a  n  d  3
Substituting into n1 x  n2 y  n3 z  d  0 gives:
2 x  y  3z  3  0
2
 
… and the vector 1 will be
 
3
 
…the value of d locates the plane.
perpendicular to each plane…
(ii) You need to verify that 2 x  y  3z  3  0 when x  2, y  4and z  1
So: 2  2  4  3  (1)  3  4  4  3  3  0 as required.
For further practice in examples like the one above, use the interactive
questions Vector plane equation using the normal.
The equation of a plane through three points
You need three points to define a plane.
A plane through the points A, B and C can be defined by the vector equation:
r  OA  λ AB  μ AC
You can get to any
point on a plane by…
…moving μ steps in
the direction AC …
…moving to a point
on the plane...
…moving λ steps in
the direction
AB …
Example 3
Find the equation of the plane through A(-2, 3, 1), B(2, 0, -4) and C(-1, 3, 2).
Solution
 2 
 
OA   3 
1
 
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 2   2   4 
     
AB   0    3    3 
 4   1   5 
     
 1  2   1 
     
AC   3    3    0 
 2   1  1
     
Substituting these into the general vector equation of a plane r  OA  λ AB  μ AC :
 2 
 4
1
 
 
 
r   3   λ  3   μ  0 
1
 5 
1
 
 
 
The following example shows you how to convert the vector equation to a
cartesian equation.
Example 4
 2 
 4
1
 
 
 
Find the cartesian equation of the plane r   3   λ  3   μ  0 
1
 5 
1
 
 
 
Solution
Step 1: Read across the vector equation:
 x   2 
 4
1
   
 
 
r   y    3   λ  3   μ  0 
z  1 
 5 
1
   
 
 
So:
x  2  4 λ  μ
y  3  3λ
z  1  5λ  μ



Step 2: Now eliminate the parameters λ and μ between the 3 equations.
y  3  3λ  λ 
From 
-

x
z
3 y
3

 2  4 λ  μ
 1  5λ  μ
x  z  3  9 λ
So:
x  z  3  9λ

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Substitute  into :
3 y 
x  z  3  9 
  x  z  3  3(3  y )
 3 
Simplifying:
x  z  3  9  3 y
 x  3y  z  6
Check: Check the points A(-2, 3, 1), B(2, 0, -4) and C(-1, 3, 2) lie on the plane:
x  3y  z  6
 A :  2  3  3  1  2  9  1  6
B : 2  3  0  (4)  2  0  4  6
C :  1  3  3  2  1  9  2  6
In practice the Cartesian equation of the plane is usually easier to use.
Finding the intersection of a line and a plane
You can find the point of intersection of a line and a plane by substituting the
vector equation of the line into the vector equation of the plane, and solving to
find the value of  (unless the line is parallel to the plane, in which case there
is no value of , or the line lies in the plane, in which case there are an infinite
number of possible values of ). You can then substitute this value of  into
the equation of the line to find the coordinates of the point of intersection.
The following example shows you how to find the intersection of a line and a
plane.
Example 5
2
 1 
 
 
Find the point of intersection of the line r   3   λ  2  and the plane
4
 3 
 
 
2x  3 y  z  6
Solution
The general point on the line is:
 x  2 
 1 
   
 
r   y    3   λ  2 
   
 
z  4 
 3 
So reading across:
x  2 λ
y  3  2 λ
z  4  3λ
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Substitute these into the equation of the plane 2 x  3 y  z  6 :
2(2  λ)  3(3  2 λ)  (4  3λ)  6
Simplifying:
4  2 λ  9  6 λ  4  3λ  6
 17  11λ  6
 11λ  11
 λ 1
Now substitute λ  1 into the equation of the line to find the position vector of the
point of intersection.
 x   2   1   1 
       
r   y    3   1 2    1
       
 z   4   3   1 
So the coordinates of the point of intersection are:
(1, -1, 1)
Check this point lies on the plane 2 x  3 y  z  6 :
2  1  3  (1)  1  2  3  1  6 as required..
Finding the distance of a point from a plane
The shortest distance of a point A to a plane is the distance AP where AP is a
line perpendicular to the plane and P is a point on the plane.
To do this:
Step 1: Find the equation of the line through A perpendicular to the
plane
Step 2: Find the point of intersection, P, of the line and the plane
Step 3: Find the distance AP.
The following example shows you how to do this.
Example 6
Find the distance of the point A(-1, 2, 4) from the plane 3x  2 y  z  11 .
Solution
 3 
 
Step 1: The direction vector perpendicular to the plane is:  2 
 1 
 
So the equation of the line through A(-1, 2, 4) is:
 1
 3
 
 
r   2   λ  2 
4
1
 
 
Step 2: The general point on the line is:
The coefficients of x, y and z
in the equation of the plane.
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 x   1
 3
   
 
r   y    2   λ  2 
z  4 
1
   
 
So reading across:
x  1  3 λ
y  2  2λ
z  4 λ
Substitute these into the equation of the plane 3x  2 y  z  11 :
3(1  3λ)  2(2  2 λ)  (4  λ)  11
Simplifying:
3  9 λ  4  4 λ  4  λ  11
 3  14 λ  11
 14 λ  14
 λ 1
Now substitute λ  1 into the equation of the line to find the position vector
of the point of intersection.
 x   1  3   2 
       
r   y    2   1 2    0 
 z   4   1  5
       
So the coordinates of the point of intersection are: P(2, 0, 5)
 2   1  3 
     
Step 3: AP   0    2    2 
5  4   1 
     
AP  32  (2) 2  12  9  4  1  14
So the distance of the point A to the plane is 14 units.
This method of finding the distance of a point from a plane can be generalised
to give the formula below:
The distance of a point (x1, y1, z1) from a plane ax + by + cz + d = 0 is given by
the formula
ax1  by1  cz1  d
a 2  b2  c2
This formula is given in your formula book, but it is easy to learn.
Example 7 shows this formula being applied.
Example 7
Find the distance of the point (1, -2, 4) from the plane 2 x  y  3z  4 .
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Solution
The equation of the plane can be written as 2 x  y  3z  4  0 , so a = 2, b = 1, c = -3
and d = -4.
 2  2   1 2    3  4   4
Distance of point from plane 
22  12  32
14

14
 14
The angle between a line and a plane
The angle between a line and a plane is the complement of the angle
between the line and the normal to the plane (i.e. the angle between a line
and a plane = 90° - the angle between the line and the normal to the plane).
You can find the cosine of the angle between the normal and the line by using
the scalar product:
For the line r  a  b and the plane r.n  d , the acute angle  between the
direction vector of the line, b, and the normal vector to the plane, n, is given
by
cos  
b.n
b n
and so the acute angle  between the line and the plane is given by
b.n
b n
sin  
The angle between two planes
The acute angle between two planes is equal to the acute angle between the
normals to the planes. This means that the angle between two planes can be
found quite easily using the scalar product.
cos  
n1.n 2
n1 n 2
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