IB Mathematics HL
What’s Your Position On Vectors?
Position Vectors and Base Vectors

 x
Every point P(x,y) in 2 has a corresponding position vector p  OP    , where O is the Origin.
 y
P(x,y)
p
O
The points (1,0) and (0,1) have position vectors i and j respectively. A linear combination of these
base vectors will generate any other vector.
 x
e.g. p     xi  yj
 y
k
Q1: Prove that if p   , then p  ki  lj
l 

Given any two points A(x1, y1) and B(x2, y2) we can travel from A to B on the vector AB . We can




 from A to B through O (or any other third point); AB  AO OB .
also travel




Now with OA  a, OB  b and AO   OA , we get





AB  AO OB  OB OA  b  a
That is,

 x 2   x1   x 2  x 1 
AB        

 y 2   y1   y 2  y1 
E.g. A(5, -3) and B(7,4)

 7  5   2
AB         
 4   3  7
1 
3
Q2: Write vectors a    and b    in i, j notation and calculate 3a – 2b giving your answer in
2
1 
i, j notation.
5 3 1

Q3: Find numbers
s and 
t such that s   t     . Illustrate your answer geometrically.
4 2 2
Q4: If p = 4i + j, q = 6i – 5j and r = 3i + 4j, find the numbers s and t such that sp + tq = r. Illustrate
your answer geometrically.

Q5: The points A and B have coordinates (3, 1) and (1, 2) respectively. Plot on squared paper the
points C, D, …, H defined by the following vector equations, and state their coordinates.
(a) c = 3a
(b) d = -b
(c) e = a – b
(d) f = b – 3a
(e) g = b + 3a
(f) h = ½ (b + 3a)
Geometric Applications
Consider the triangle OAB with P, Q mid-points of OA and OB, respectively.
A
P
B
Q
O
1
a
2
We can let O represent the origin and use vectors. We then have:
1
q b
2
p


Now we can determine AB and PQ in terms of a and b.

AB  b  a

PQ  q  p
1
1
b a
2
2
1
 b  a
2



Since PQ is a multiple of AB we can deduce that PQ // AB and is half as long. We have used vector
methods to prove a well-known geometric property about triangles and their mid-segments.
Q6: Prove that the midpoint of the line segment from Ax1, y1 to Bx 2, y 2  is
x1  x 2 y1  y 2 
,
 2
2 
Q7: Points A and B have coordinates (2, 7) and (-3, -3) respectively. Use a vector method to find the
coordinates of C and D, where



(a) C is the point such that AC  3AB
(b) D is the point such that AD  35 AB
Q8: C is the point on AB such that AC : CB = 4 : 3. Express c in terms of a and b.


Q9: If C is the point on AB such that AC  t AB , prove that c = tb + (1-t)a.
Q10: In the figure below, OABC is a quadrilateral and P, Q, R and S are the mid-points of the sides
OA, AB, BC and CO, respectively.
Given that OA = a, OB = b and OC = c, express the following

vectors in terms of a, b and c:
(a) PS , (b) AC , (c) QR
What can you deduce about the lines 
PS, AC and
 QR? 

O


B
Q11: In the previous question, X is the mid-point of PR, and Y is the mid-point of QS. Express OX
and OY in terms of a, b and c. State clearly in words the deduction which can be made from these
expressions.


Scalar (Dot) Product
Vector multiplication can result in another vector (Cross Product – done later in HL) or in a scalar
(Dot Product). The dot product of two vectors, v and w, is defined as the product of the length of w
and the length of the projection of v onto w (or vice versa).
V
v

O
w
P
W
OP = the length of the projection of v onto w
= v cos
So,
v  w  w  v cos
Note that if 90 <  < 180 the projection is in the opposite sense, so that a dot product may be
negative.
Some special properties of the dot product become evident from this definition.


v  v  v since  = 0
if vw then v  w  0 since  = 90
2
ii  i  1
2

in particular,
2
j j  j  1
and i  j  0
 x1 
 x2 
Using this last property with v     x1i  y1 j and w     x 2 i  y 2 j , we get
 y1 
 y2 
v  w   x1i  y1 j x 2 i  y 2 j
 x1 x 2 i  i  x1 y 2 i  j  x 2 y1 j  i  y1 y 2 j  j
= x1 x 2  y1 y 2
Thus we have two definitions of dot product.
v  w  v w cos
and
v  w  x1 x 2  y1 y 2
Now, given two vectors in component form it is a fairly simple matter to find , the angle between
the vectors.
E.g.
 4
 3
v    and w   
  1
  2
v  17 , and w  13
and v  w  4(3)  ( 1)( 2)  14
 14  17 13 cos
or
cos  0.9417
  19.7
Dot product is used in Physics to determine Work done in applying a force F to displace an object
along a vector d.
W  Fd
Question: Dot product can be negative. What about work?!
Dot product can also be used to show some interesting geometric properties
Consider;
c
a

b
c  a b
c  c  (a  b)  (a  b)
= a  a  2a  b  b  b
c  a  b  2 a b cos
2
Look familiar?
2
2
 


3
4
1 
Q12: Let a =  , b =   and c =  . Calculate a b, a c and a (b + c), and verify that a (b +
2
2 
4
c) = a b + a c.


x 
x 
Let r1  1 and r2  2 . Calculate r2  r1 and interpret your result geometrically.
Q13:
y1
y 2 
 
  

Q14: Find the angle between the line joining (1, 2) and (3, -5) and the line joining (2, -3) to (1, 4).


 vectors a and b such that a  b , and complete the parallelogram OACB. Mark on
Q15: Draw two
your diagram the vectors a – b and a + b. By considering the scalar product a  b a  b , prove
that the diagonals of a rhombus intersect at right angles.
Vectors in three dimensions


The power of vector methods is best appreciated when they are used to do geometry in three
dimensions. As you already know, this means that you work on a coordinate system with x-, y-, and zaxes. Looking at the familiar x- and y-axis on a plane (your sheet of paper), you can imagine the zaxis coming straight up at you, at right angles to both the x- and y- axes.
A vector p in three dimensions is a translation of the whole space relative to a fixed coordinate
l 
framework. It is written as m, which is a translation of l , m and n units in the x-, y-, and z 
n 
1 
0
directions respectively. It can also be written in the form li + mj + nk, where i  0, j  1 ,
 
 

0
0
0
k  0 are basic unit vectors in the x-, y-, and z- directions.
 


1 
Q16: What is the distance between two points x1, y1,z1 and x2, y2,z2  in space?

1 
3


Q17: Write the vectors a  2 andb  1  in i,j, k notation, and calculate 3a – 2b giving your
 
 
1 
2 
answer in i, j, k notation.
Q18: If v  i  8j  k and w  2i  3j  k , find 2v – 3w giving your answer in column or component


form.


Q19: Looking back at what we know so far about vectors in two dimensions, how would the scalar
product of vectors be different for three dimensions? Write down the vector scalar product in three
dimensions, using both definitions.
3 
1
33 
Q20: Let u  1, v  9 and w  13. Calculate u v , u w and v w. What can you deduce
 
 
 
4 
3 
28
about the vectors u, v and w?
 

Q21: Which of the following vectors are perpendicular to each other?



(a) 2i  3j  6k

  
(b) 2i  3j  6k
(c) 3i  6j  2k
(d) 6i  2j  3k
1 
 j  2k .
Q22: Find unit 
vectors in the same direction
as 2 and 2i 

 
2 
Q23: Find the angle between the line joining (1,
3, -2) and (2, 5, -1) and the line joining (-1, 4, 3) and
(3, 2, 1).

Q24: Two aeroplanes are flying in directions given by the vectors 300i  400j  2k and
100i  500j  k . A person from the flight control centre is plotting their paths on a map. Find the
acute angle between their paths on the map.


Distance from a point to a line
Vectors can be used to find the distance from a point to a line. Here’s one method.
Consider line L: 2x - 3y + 6 = 0 and the point Q(1,7)
Q
L

P
B
A

 3
First, determine two points on L, say the intercepts A(-3, 0) and B(0, 2). Then the vector AB   
 2
generates L. Let P(x, y) be some point on L.
 x 1   x 1 


 x   1  x  1
2
  2

Now QP        
 . But, y  x  2 so QP   2
  x  5
x

2

7
3
 y  7  y  7
3
 3



The distance from L to Q is the perpendicular distance. Thus we want AB QP  0 .
 3  x  1 
    2 x  5  0
 2  3

4
3x  3  x  10  0
3
13
x  13
3
x3
so y  4

Name this point C(3, 4) and CQ will represent the desired distance.

  2
CQ  q  c     ( 2) 2  32  13
 3
Another method leads to a familiar formula.
Consider the same line, L, and the point Q(1, 7)
Q

B
L
P
A

 3
We have already seen that AB    defines the direction of the line L and that B is the point (0, 2).
 2

  2
The normal, n, to the line is the vector perpendicular to AB or n    . If we choose the normal
 3 
that contains Q and P then the distance QP will be the length of the projection of

 1   0   1
v  BQ          onto the normal vector through Q as shown below.
 7   2   5
n
Q
B
 P
A
We noted earlier that the length of the projection of a vector, v, onto another vector, n, is given by
n
(since v  n  v  n cos )
v cos  v 
n
n
(Note:
 n is a unit normal)
n
Thus, in this case,
v cos  QP  distance from Q to L.
v cos  v 
n
n
 1   2 1
      
 5   3  13
1
 (2  15) 
13
13

 13
13
Generalizing this method to the line L: ax + by + c = 0 and the point Q(x1, y1) leads to the formula for
the distance from a point (x1, y1) to a line L;
d=
ax1  by1  c
a2  b2
EXERCISE?
The vector equation of a line
In order to get to any point on a given road we must first get to the road and then walk some distance
along the road. We can use the same concept in the co-ordinate plane to generate the general vector
 x
r =   , where (x, y) is on a given line L. Consider the diagram below.
 y
L
P
r
O
R
 0
 2
Suppose P is the point (0,1) with position vector p =   and the direction of L is given by d =   .
 1
 1
Now, to get to R from O we first travel to P and then along L some scalar multiple of d. That is,
r = p + td
or,
 x  0  2
      t 
 y  1  1
These are vector equations for the line L. We can easily convert L into Cartesian form by eliminating
the parameter t.
1
x
2
y=1+t  t=y–1
1
1
 y  1  x or y = x + 1 or x - 2y = -2
2
2
  1
We can also eliminate t by using a normal vector, n, (say n =   ) and taking the dot product with
 2
r.
x = 0 + 2t  t =
That is,
r·n = p·n + td·n
but d·n = 0
 x    1  0   1
therefore           
 y  2   1  2 
or
x - 2y = -2
As an exercise you may want to confirm that, if a unit normal n is used then the resulting value p  n
equals the distance from the line to the origin!
To recap:
Points on a line through A in the direction of b have position vectors r  a  b ,
where  is a variable scalar. This is called the vector equation of a line.


Q25: Write down vector equations for the line through the given point in the specified direction.
Then eliminate the parameter to obtain the Cartesian equation.
1
(a) (2, -3),  
2
(b) (5, 7), parallel to the x-axis
2 
(c) (0, 0),  
1
Q26: Find vector equations for lines with the following Cartesian equations.
 x = 2
(a)

(b) 2x – 5y = 3
Q27: Find the coordinates of the points common to the following pairs of lines, if any.
5 1
3  1
(a) r        , r       
1 2 
1 1

2  1 
4 2
(b) r        , r      
1 3
0 6 
3 2 
Q28: Find the coordinates of the point where the line with vector equation r       
4  1



intersects the line with Cartesian equation 2x + y = 7.
Q29: Find the vector in the direction of line l with Cartesian equation 3x – y = 8. Write down a vector

equation for the line through P(1, 5) which is perpendicular to l. Hence
find the coordinates of the
foot of the perpendicular P to l.
Q30: Two airliners take off simultaneously from different airports. As they climb, their positions
relative to an air traffic control centre t minutes later are given by the vectors
5  8 
13  6 




and r2  26  t 3 , the units being kilometers. Find the coordinates of the
r1  30  t 2
   
   
0  0.5
0  0.6
point on the ground over which both airliners pass. Also find the difference in heights, and the
difference in the times, when they pass over that point.

Q31: The line 
l is given by r  i  k   i  2j  3k. Find
(a) the distance of the point P(1, -9, 9) from l,
(b) the coordinates of the reflection of P in l.

Cartesian equation of a line in three dimensions
You can find the Cartesian equation of a line in three dimensions just as the equation of a line in two
dimensions, by eliminating the parameter. Try this with the following vector, then generalize.
x 5 1 
y  3    2 
     
z  4  1

NOTE:
 There are two equations, not a single equation. You will see later that the equation of the form
ax + by + cz = d, which you might have expected to be the equation of a line in three
dimensions, is actually the equation of a ___________.
 You can avoid the Cartesian form by turning it immediately into vector form. Why would this
be useful? Give an example of how to do this.
Q32: Find the Cartesian equations of the line L joining (-1, 4, 1) to (3, 6, 2). Does this line intersect
the straight line M with vector equation r  i  j  k  i  2j  2k?
x3
 y  7  1 z .
Q33: The Cartesian equation of the line L is
2
 8, 0) lies on line L.
(a) Show that the point P(5,
(b) Find the angle in radians between L and PQ where Q is the point (2, 1, 3).

Motion and Vectors
Consider the following example of one-dimensional motion. An object moves along a line with its
position determined by its distance from some fixed reference point, O. The object starts at the point
A, 5 metres right of O, and then travels at 3ms-1. Where is the object after t seconds?
O
A
Clearly, its position will be at 5 + 3t metres from O, assuming it travels to the right.
Now we add a dimension. Consider the object moving along a line in  2 , with velocity vector
 3 ms 1 
 , passing through the point P(-8,5).
v
  4 ms 1 
y
 P(-8,5)
x
v
Following the one-dimensional example, we can determine the position of the object by the vector
 x    8  3 
equation       t   , where t is the time elapsed. Further, we are moving 3 ms-1
 y  5    4
horizontally and 4 ms-1 down. The net effect is the magnitude of the velocity vector, or the speed.
Thus, the speed is v  32  ( 4) 2  5 ms 1 .
We can now treat linear motion problems in two dimensions using equations such as
 x   a   v1 
      t 
 y  b   v 2 
This is similar to the vector equation for a line, except that t represents time travelled. Note that this
idea is easily transferred to three (or more!) dimensions.
A different explanation
Suppose that in the vector equation of a line, r = a +tp, in either two or three dimensions, the scalar t
represents time. Let P be the point whose position vector is r. How can you calculate how fast P is
moving along the line?
Example:
A chair in a chair lift has position vector r = (2i + 3j + k) +t(2i – 2j + k), where t is the time in
minutes and unit vectors are in meters. Find its velocity and its speed.
The position P of the chair at time t is given by r = (2i + 3j + k) +t(2i – 2j + k).
At time t  t , the position vector is given by r  r , where
r  r  2i  3j  k  t  t2i  2j  k.
By subtraction



r  2i  3j  k  t  t2i  2j  k 2i  3j  k t2i  2j  k

 t2i  2j  k
r
 2i  2j  k
So
t
 small, the average velocity will be very close to the actual velocity at P, which is the
If t is very
limit of the average velocity as t tend to 0. That is,

r dr
v  lim

t0 t
dt
r
In this case,
is a constant,
so v  2i  2j  k .

t

The speed is the magnitude of the velocity, and so is given by
Speed v

 2i  2j  k
 2 2  2  12
2

 93
The speed of the chair is 3 meters per minute.

If a point is moving with position vector r = a +tp, the velocity, in
appropriate units, is p and the speed is p .
Q34: Investigate whether or not it is possible to find numbers s and t which satisfy the following
vector equations. If it is, find them.

3
2
0
     
1  5 1 
(a) s4  t 1  11
(b) s2   t 1  7
     
     
1  0  2 
3 1 11 
Q35: Find if the following pairs of lines intersect. If they do, give the point of intersection.

(a) r = 3i – 2j – k + s(-2i + 3j - 5k)and r = 8i + 4j + k + t(-i – 3j + k)
(b) r = -3i + 5j + 2k + s(-i - 2j - 4k) and r = 8i + 5j + k + t(-i – 3j + 3k)
Q36: If p = 2i – j + 3k, q = 5i +2j and r = 4i + j + k, find a set of numbers f, g and h such that
fp + gq + hr = 0. What does this tell you about the translations represented by p, q and r?
Q37: For each of the following sets of points A, B, C and D, determine whether the lines AB and CD
are parallel, intersect each other, or are skew.
(a) A(3, 2, 4), B(-3, -7, -8), C(0, 1, 3), D(-2, 5, 9)
(b) A(-5, -4, -3), B(5, 1, 2), C(-1, -3, 0), D(8, 0, 6)
(c) A(2, 0, 3), B(-1, 2, 1), C(4, -1, 5), D(10, -5, 1)
Q38: Find the point of intersection of the following pairs of lines.

(a)
x 1 y  2 z 1
x3 y2 z




and
3
1
2
4
2
3
(b)
x2
1 z
x 1 y  4
y

 z 1
and
4
2
3
3

Q39: Find the coordinates of the foot of the perpendicular from (3, 9, -2) to the line with vector
 equation r  3i  j  i  j  k . Find also the distance between (3, 9, -2) and its reflection in the


line.
Q40: A student displays her birthday cards on strings which she has pinned to opposite walls of her

room, whose floor measures 3m by 4m. Relative to one corner of the room, the coordinates of the
ends of the first string are (0, 3.3, 2.4) and (3, 1.3, 1.9) in meter units. The coordinates of the ends of
the second string are (0.7, 0, 2.3) and (1.5, 4, 1.5). Find the difference in the heights of the two strings
where one passes over the other.
Q41: A balloon flying over flat desert land reports its position at 7:40am as (7.8, 5.4, 1.2), the
coordinates being in kilometers relative to a checkpoint on the ground. By 7:50am its position has
changed to (9.3, 4.4, 0.7). Find its speed and, assuming that it continues to descend at the same speed
along the same line, find the coordinates of the point where it would be expected to land, and the time
when this would occur.
Work Chapter 24 and 25 from Cambridge.
Keep adding in problems from Stewart or other.
Sources:
Pure Mathematics Book 1, Fourth Edition, Longman, J.K. Backhouse, S.P.T. Houldsworth, and P.J.F.
Horril
Mathematics for the IB Diploma HL2, Cambridge, Hugh Neill and Douglas Quadling
Note: Transfer all of above to 3-D vectors as well.
Last Words
It is important when teaching vectors to focus on vector methods. Encourage students to use the new
methods rather than fall back on coordinate geometry techniques. They will follow your lead. Help
them enjoy a whole other mathematical world!