Mathematics

Key unit competence: By the end of this unit, learners should be able
to apply properties of collinearity and orthogonarity to solve problems involving
vectors.

Unit outline

• Collinear points

• Orthogonal vectors

Introduction

Unit Focus Activity

1. Consider the points
(a) P(–1, 2) Q(1, 4) and R(3, 6)

(b) L(3, –2) Q(1, 0) and N(0, 3)

(i) W i t h o u t p l o t t i n g determine whether the
    three points in each are collinear (lie on the same straight line)

(ii) State one advantage of three or more objects lying
  on the same straight line.
 2. Consider the column vectors.

Determine whether the two pairs of vectors are orthogonal in each case.

Determine whether the two pairs of vectors are orthogonal in each case.

10.1 Collinear points

10.1.1 Definition of colinearity

Activity 10.1

1. You are given three points A (2, 2), B (3, 3) and C (6, 6).

(a) Plot the points ABC on a Cartesian plane.

(b) Join the points A, B and C using a ruler.

2. Given the points P (–2, 3), Q (3, 3) and R (1, –5).

(a) Plot the points P, Q and R on a Cartesian plane

(b) Join the points P, Q and R using a ruler

3. What observation do you make in the plotted points A, B, C and P, Q, R?

4. How would you have determined what you observed in step 3 without
    drawing?

Three or more points are said to be collinear if they lie on a single straight line.
For example the points A, B and C in Fig. 10.1 are collinear because they lie on a
single straight line.

Points E, F and G in Fig.10.2 below are not collinear because they don’t lie on the
same straight line.

10.1.2 Verifying collinearlity of points using vector laws A B C

Activity 10.2

Observe the straight line PQR in graph 10.1 below.

1. Determine the column vectors PQ and QR.

2. Express vector PQ in terms of QR.

3. Since points P, Q and R are collinear, state one condition
   for collinearlity guided by relationship of vectors PQ and
   QR that you obtained in step 3.

Consider another straight line ABC in graph 10.2 below with point B located
 between A and C as shown.

If AB = kBC where k is a scalar, then
 AB is parallel to BC. Since B is a common point between
 vectors AB and BC, then A, B and C lie
 on a straight line, i.e. the points A, B and
 C are collinear.

Example 10.1

Show that the points A (0, -2), B (2, 4) and C (-1, -5) are collinear.

(i) 2 = 3k ⇒ k = –2/3

(ii) 6 = –9k ⇒ k = –6/9 = –2/3

Since the value of k is the same for the two cases (i) and (ii), i.e. AB= –2/3BC, and B
is a common point of two vectors AB and BC, then points A, B and C are collinear.

Example 10.2

For what value of k are the following points collinear? A(1, 5), B(k, 1) and
C(11, 7)

Solution:

Let the points be A, B and C. For the points to be collinear, B can be a common
point and therefore we get; AB = aBC where a is a scalar

Exercise 10.1

1. Verify whether the following points are collinear.

(a) A(2, 3), B(9, 8) and (5, 4)

(b) P(–1, 1), Q(5, 1) and R(2, 0)

(c) X(–2, 3), Y(7, 0) and Z(1, 2)

(d) R(1, 2), S(4, 0) and T(–2, 4)

2. (a) The following points are collinear, find the values of
    unknown in each case.
(i) (1, 0), (k, 3), (4, 5)

(ii) (5, 1), (x, 2), (x, 7)

(iii) (2, 4), (1, k), (6, k)

(b) Plot all the points in (a) on a Cartesian plane. What do you
     observe?

10.1.3 Applications of collinearlity in proportional division of lines A B C

        Activity 10.3

AP: PB = 4: –1.
The negative sign is required as PB is in the opposite direction to AB. We
say that P divides AB externally in the ratio 4: –1, or P divides BA in the
ratio –1 : 4

Whenever P is outside AB (on either side), AP: PB will be negative, and
we say that P divides AB externally.

Exercise 10.2

1. In the diagram in Fig.10.9, OD = 2OA, OE = 4OB, OA = a
   and OB = b. Points B, A and C are collinear.

(a) Express the following vectors in terms of a and b. OD, OE, BA
     and ED.

(b) Given that BC = 3BA, express:

(i) OC

(ii) EC, in terms of a and b.

(c) Hence show that the points E, D and C lie on a straight line.

2. In the diagram in Fig 10.10, OA = a, OB = b and M is the mid-point of
    AB. Points A, M and B are collinear.

Find OM in terms of a and b.

3. In Figure 10.11, OABC is a parallelogram, M is the mid-point of
    OA and AX = 7AC, OA = a and OC = c. Points A, Y and B, M, X
     and Y, A, X and C are collinear.

(a) Express the following in terms ofa and c.

(i) MA (ii) AB

(iii) AC (iv) AX

(b) Using triangle MAX, express
    MX in terms of a and c.

(c) If AY = pAB, use triangle MAY
to express MY in terms of a, c
and p.

d) Also if MY = qMX, use the result in (b) to express MY in terms of
        a, c and q.
(e) Hence find p and q and the ratio
      AY: YB
4. ORST is a parallelogram and S, P and T are collinear in Fig.10.12.

If ST = 4SP, express the following in
    terms of r and or t.
(a) RS        (b) ST       (c) SP

(d) RP         (e) OP
5. In Fig. 10.13, PQR is a triangle,
     PQ = a, PR = b and S is the midpointof RQ.

Express the following in terms of aand b.
(a) OY, OX, AB, and XY.

(b) Given that AC = 6AB, express
     OC and XC in terms of a and b.

(c) Use the results for XY and XC to
      show that points X, Y and C lie
       on a straight line.

7. In the diagram below OA = a and
    OB = b, M is the mid-point of OA and
    P lies on AB such that AP = 3 AB.

(a) Express the following in terms of a and b: AB, AP, MA and MP

(b) If X lies on OB produced such  that B = BX, express MX in
     terms of a and b.

(c) Show that MPX is a straight line.

10.2 Orthogonal vectors

  Activity 10.4

1. Research from reference books or internet how to determine the
     product of two column vectors.

2. You are given the vectors

a) Find the product of a and b.

b) Plot the vectors in a Cartesian plane. What is the angle
    between them?

3. You are given the vectors

(a) Find the product of p and q.

(b) Plot the vectors on a Cartesian plane. What is the
     angle between them?

4. Based on your results in 2 and 3, what is the relationship between
    two perpendicular vectors and the product of the vectors?

Two vectors are said to be perpendicular if
the angle between them is 90° (i.e. if they form a right angle).

The product of two vectors must be zero
for the angle between them to be 90º. Consider two vectors

and

Two vectors are said to be orthogonal if the sum of product of tops and the product
of bottoms is zero. If vectors a and b are orthogonal, then
x₁x₂ + y₁y₂ = 0.

Note:

Example 10.7

(a) Show that the vectorsandare orthogonal vectors

Solution

(a) Vectors a and b are orthogonal
    vectors if x1x2 + y1y2 = 0
                 = (2 × – 1) + (1 × 2)
                 = –2 + 2 = 0.

(b) These vectors are orthogonal.

Exercise 10.3

1. Show whether the following vectors are orthogonal.

2. Consider points A(5, 3), B(2, –1) and C(7, –3,). Find:

(i) BA and BC

(ii) Show whether BA and BC are orthogonal. Give reasons.

3. Letand be two
    vectors. Find the values of k if; a and b are perpendicular

4. Are the following given vectors perpendicular? Write true/false for
    the result obtained.

5. The co-ordinates of P, Q, R and S are (0, 1), (7, 8), (1, –1) and (9, 7).

(a) Find PQ and RS.

(b) What is the relationship between PQ and RS?

Unit Summary

• Collinearlity: Three or more points are said to be collinear if they lie on a
   single straight line. For example if A, B and C are three
    points on the same straight line ABC, then AB = kBC where B becomes a
    common point.

• Collinear points find their applications in proportional line
   divisions.  For example, consider the following cases;

AP: PB = 3: l, then we say that P divides AB internally in the ratio 3: 1.
(P lies between A and B). Note: The direction is important as
P divides BA in the ratio 1: 3.
AP = 3/4 AB and PB = 1/4 AB

• Orthogonal vectors: Two vectors are said to be perpendicular if the
 angle between them is 90° (i.e. if they form a right angle).

The product of two vectors must be zero for the angle between them to be
 90º.
Consider two vectorsand

the sum of the product of tops and bottoms is defined by the
formula;
        = x₁x₂ + y₁y₂
If vectors a and b are orthogonal,
then x₁x₂ + y₁y₂ = 0.

Unit 10 Test