# Vectors II - Mathematics Form 3 Notes

## Vectors in 3 Dimensions

• 3 dimensional vectors can be represented on a set of 3 axes at right angles to each other (orthogonal), as shown in the diagram.
• Note that the z axis is the vertical axis.
• To get from A to B you would move:
4 units in the x-direction, (x-component)
3 units in the y-direction, (y-component)
2 units in the z-direction. (z-component)

In component form:

• In general:

## Column and Position Vectors

• In three dimensions, a displacement is represented b a column vector of the form  where p,q and r are the changes in x,y,z directions respectively.

Example

The displacement from A (3, 1, 4 ) to B ( 7 ,2,6) is represented b the column vector,

The position vector of A written as OA is where O is the origin

Addition of vectors in three dimensions is done in the same way as that in two dimensions.

Example

### Column Vectors in Terms of Unit Vectors

In three dimension the unit vector in the x axis direction is = ,that in the dirction of the y axis is = while that in the direction of z – axis is =

#### Diagrammatic Representation of the Vectors.

Three unit vectors are written as;

Example

Express vector in terms of the unit vector i, j and k

Solution

Note;

• The column vector  can be expressed as ai + bj + ck

## Magnitude of a 3 Dimensional Vector.

• Given the vector AB = xi + yj + 2k,then the magnitude of AB is written as |AB| = √(x2 +y2 +z2)

• This is the length of the vector.
• Use Pythagoras’ Theorem in 3 dimensions.
AB2 = AR2 + BR2
= (AP2 + PR2) + BR2
=(xB − xA)2 + (yB − yA)2 + (zB − zA)2
→
and if
u = AB then the magnitude of u, |u| = length of AB

## Distance Formula for a 3 Dimensionsional Vector

• Recall that since: = , then  if then |u| = √(x2 +y+z2)
• Since x = xB  xA and y = yB yA and zB  zA

Example:

1. If A is (1, 3, 2) and B is (5, 6, 4)
→
Find |AB|
2. If . Find |u|

Solution

## Parallel Vectors and Collinearity

### Parallel Vectors

• Two vectors are parallel if one is scalar multiple of the other.i.e vector a is a scalar multiple of b ,i.e. a =kb then the two vectors are parallel.

Note;

• Scalar multiplication is simply multiplication of a regular number by an entry in the vector

### Multiplying by a Scalar

• A vector can be multiplied by a number (scalar).e.g. multiply a by 3 is written as 3a.
• Vector 3a has three times the length but is in the same direction as a .In column form, each component will be multiplied by 3.
• We can also take a common factor out of a vector in component form. If a vector is a scalar multiple of another vector, then the two vectors are parallel, and differ only in magnitude. This is a useful test to see if lines are parallel.

Example

if

### Collinear Points

Points are collinear if one straight line passes through all the points. For three points A, B, C - if the line AB is parallel to BC, since B is common to both lines, A, B and C are collinear.

#### Test for collinearity

Example

A is (0, 1, 2), B is (1, 3, –1 ) and C is (3, 7, –7) Show that A, B and C are collinear.

Solution

→        →
AB and BC are scalar multiples, so AB is parallel to BC. Since B is a common point, then A, B and C are collinear.

In general the test of collinearity of three points consists of two parts

• Showing that the column vectors between any two of the points are parallel
• Showing that they have a point in common.

Example

A (0,3), B (1 ,5) and C ( 4,1 1 ) are three given points. Show that they are collinear.

Solution

AB and BC are parallel if AB = kBC, where k is a scalar

Therefore AB//BC and point B (1, 5) is common. Therefore A,B,and C are collinear.

Example

Show that the points A (1, 3, 5) ,B(4, 12, 20) and C are collinear.

Solution

Consider vectors AB and AC

Hence k =2/3
AC =2/3 AB
Therefore AB//AC and the two vectors share a common point A.The three points are thus collinear.

Example

In the figure above OA = a OB = b and OC = 3OB

1. Express AB and AC in terms of a and b
2. Given that AM = ¾ AB and AN = ½ AC, Express OM and O in terms of a and b
3. Hence ,show that OM and N are collinear

Solution

1. AB = OA + OB
= − a + b
AC = − a + 3b
2. OM =OA + AM
= OA + ¾AB
= a +¾(− a + b)
= a − ¾ab
= ¼a + ¾b
ON =OA +AN
=OA + ½AC
a + ½(− a + 3b)
a = a − ½a + 3/2b
= ½ a + 3/2b
3. OM = kON ¾ b + ¼a = k/2a + 3k/2b
Comparing the coefficients of a;
1/4 = k/2
k = 1/2
Thus, OM = 1/2 ON.
Thus two vectors also share a common point ,O .Hence, the points are collinear.

## Proportional Division of a Line

### Internal Division

• In the figure below, the line is divided into 7 equal parts

• The point R lies 4/7 of the ways along PQ if we take the direction from P to Q to be positive, we say R divides PQ internally in the ratio 4 : 3.
• If Q to P is taken as positive,then R divides QP internally in the ratio 3 : 4 .Hence, QR : RP = 3 : 4 or, 4 QR = 3RP.

### External Division

• In internal division we look at the point within a given interval while in external division we look at points outside a given interval,
• In the figure below point P is produced on AB

• The line AB is divided into three equal parts with BP equal to two of these parts. If the direction from A to B is taken as positive, then the direction from P to B is negative.
• Thus AP : PB = 5 : -2.In this case we say that P divides AB externally in the ratio 5 : −2 or P divides AB in the ratio 5 : −2.

### Points, Ratios and Lines

#### Finding the ratio in which a point divides a line.

Example:

The points A(2, –3, 4), B(8, 3, 1) and C(12, 7, –1 ) form a straight line. Find the ratio in which B divides AC.

Solution

B divides AC in ratio of 3 : 2

#### Points Dividing Lines in Given Ratios.

Example:

P divides AB in the ratio 4:3. If A is (2, 1, –3) and B is (16, 15, 11 ), find the co-ordinates of P.

Solution:

→                →       →
AP = 4   so 3AP = 4PB
→     3
PB

3(p a) = 4(b p)
3
p – 3a = 4b – 4p
7p = 4b + 3a
p = 1/7(4b + 3a)

#### Points Dividing Lines in Given Ratios Externally.

Example:

Q divides MN externally in the ratio of 3:2. M is (–3, –2, –1 ) and N is (0, –5, 2).Find the co-ordinates of Q.

Note that QN is shown as –2 because the two line segments are MQ and QN, and QN is in the opposite direction to MQ.

## The Ratio Theorem

The figure below shows a point S which divides a line AB in the ratio m : n

Taking any point O as origin, we can express s in terms of a and b the positon vectors of a and b respectively.

OS = OA + AS
But AS =
AB
m+n
Therefore, OS = OA +  m   AB
m+n
Thus S = a   (−a + b)
m+n
= a −   a +  m  b
m+n     m+n

= (1 − m  )a m b
m+n       m+n
= (m+n−m)a +  m b
m+n         m+n
n   a +  m  b
m+n     m+n

This is called the ratio theorem. The theorem states that the position vectors s of a point which divides a line AB in the ratio m: n is given by the formula;

S =   n  +  m  b
m+n      m+n

where a and b are positon vectors of A and B respectively. Note that the sum ofco-ordinates n/m+n and m/m+m= 1

Thus ,in the above example if the ratio m : n = 5 : 3
Then m = 5 and n = 3
OR =
3  a + b
5+3    5+3

Thus, r = 3/8a + 3/8b

Example

A point R divides a line QR externally in the ratio 7 : 3. If q and r are position vectors of point Q and R respectively, find the position vector of p in terms of q and r.

Solution

We take any point O as the origin and join it to the points Q, R and P as shown below

QP: PR = 7: −3
Substituting m = 7 and n = −3 in the general formulae;
OP =    −
3       7    r
7+(−3)      7+(−3)

P = −3/4 q + 7/4r

Vectors can be used to determine the ratio in which a point divides two lines if they intersect

Example

In the below OA = a and OB = b. A point P divides OAin the ratio 3:1 and another point O divides AB in the ratio 2 : 5. If OQ meets BP at M

Determine:

1. OM : MQ
2. BM : MP

Let OM : MQ = k : (1 – k) and BM –MP = n : ( 1 – n)

Using the ratio theorem
OQ =
5/7 a + 2/7 b
OM= kOQ
= k
(5/7 a+ 2/7 b)
Also by ratio theorem;
OM = nOP +( 1 – n )OB
But OP =
3/4 a
Therefore, OM = n (3/4 a) + (1−n)b
Equating the two expressions;
k(5/7 a+ 2/7 b) = n (3/4a) + ( 1−n )b
Comparing the co-efficients
5/7 k = 3/4n………..1
2/7k = 1 − n……..
2
k = 21/25 and n = 10/13
The ratio BM :MP = 10/13 : 3/13
= 10: 3

## Past KCSE Questions on the Topic

1. The figure below is a right pyramid with a rectangular base ABCD and VO as the height. The vectors AD= a, AB = b and DV = v

1. Express
1. AV in terms of a and c
2. BV in terms of a, b and c
2. M is point on OV such that OM: MV=3:4, Express BM in terms of a, b and cSimplify your answer as far as possible
2. In triangle OAB, OA = a, OB = b and P lies on AB such that AP: BP = 3.5
1. Find the terms of a and b the vectors
1. AB
2. AP
3. BP
4. OP
2. Point Q is on OP such AQ = −5/8a + 9/40b
Find the ratio OQ: QP
3. The figure below shows triangle OAB in which M divides OA in the ratio 2: 3 and N divides OB in the ratio 4:1 AN and BM intersect at X

1. Given that OA = a and OB = b, express in terms of a and b:
1. AN
2. BM
2. If AX = sAN and BX = tBM, where s and t are constants, write two expressions for OX in terms of a, b, s and t
Find the value of s
Hence write OX in terms of a and b
4. The position vectors for points P and Q are 4i + 3j and 6j + 6 k respectively. Express vector PQ in terms of unit vectors i, j and k. Hence find the length of PQ, leaving your answer in simplified surd form.
5. In the figure below, vector OP = p and OR =r. Vector OS = 2r and OQ = 3/2p.

1. Express in terms of p and r (i) QR and (ii) PS
2. The lines QR and PS intersect at K such that QK = mQR and PK = nPS, where m and n are scalars. Find two distinct expressions for OK in terms of p,r,m and n. Hence find the values of m and n.
3. State the ratio PK: KS
6. Point T is the midpoint of a straight line AB. Given the position vectors of A and T are i−j + k and 2i+1½ k respectively, find the position vector of B in terms of i, j and k
7. A point R divides a line PQ internally in the ration 3:4. Another point S, divides the line PR externally in the ration 5:2. Given that PQ = 8 cm, calculate the length of RS, correct to 2 decimal places.
8. The points P, Q, R and S have position vectors 2p, 3p, r and 3r respectively, relative to an origin O. A point T divides PS internally in the ratio 1:6
1. Find, in the simplest form, the vectors OT and QT in terms p and r
1. Show that the points Q, T, and R lie on a straight line
2. Determine the ratio in which T divides QR
9. Two points P and Q have coordinates (-2, 3) and (1 , 3) respectively. A translation map point P to P’ (10, 10)
1. Find the coordinates of Q’ the image of Q under the translation
2. The position vector of P and Q in (a) above are p and q respectively given that mp – nq =
Find the value of m and n
10. Given that qi + 1/3j + 2/3 k is a unit vector, find q
11. In the diagram below, the coordinates of points A and B are (1, 6) and (15, 6) respectively). Point N is on OB such that 3 ON = 2 OB. Line OA is produced to L such that OL = 3 OA

1. Find vector LN
2. Given that a point M is on LN such that LM: MN = 3: 4, find the coordinates of M
3. If line OM is produced to T such that OM: MT = 6:1
1. Find the position vector of T
2. Show that points L, T and B are collinear
12. In the figure below, OQ = q and OR = r. Point X divides OQ in the ratio 1:2 and Y divides OR in the ratio 3:4 lines XR and YQ intersect at E.

1. Express in terms of q and r
1. XR
2. YQ
2. If XE = mXR and YE = nYQ, express OE in terms of:
1. r, q and m
2. r, q and n
3. Using the results in (b) above, find the values of m and n.
13. Vector q has a magnitude of 7 and is parallel to vector p. Given that p= 3i – j + 1½ k, express vector q in terms of i, j, and k.
14. In the figure below, OA = 3i + 3j ABD OB = 8i – j. C is a point on AB such that AC:CB 3:2, and D is a point such that OB//CD and 2OB = CD (T17)

Determine the vector DA in terms of I and j
15. In the figure below, KLMN is a trapezium in which KL is parallel to NM and KL = 3NM

Given that KN = w, NM = u and ML = v. Show that 2u = v + w
16. The points P, Q and R lie on a straight line. The position vectors of P and R are 2i + 3j + 13k and 5i – 3j + 4k respectively; Q divides SR internally in the ratio 2: 1 . Find the
1. Position vector of Q
2. Distance of Q from the origin
17. Co-ordinates of points O, P, Q and R are (0, 0), (3, 4), (11 , 6) and (8, 2) respectively. A point T is such that the vector OT, QP and QR satisfy the vector equation OT = QP + ½QT. Find the coordinates of T.
18. In the figure below OA = a, OB = b, AB = BC and OB: BD = 3:1

1. Determine
1. AB in terms of a and b
2. CD, in terms of a and b
2. If CD: DE = 1:k and OA: AE = 1:m determine
1. DE in terms of a, b and k
2. The values of k and m
19. The figure below shows a grid of equally spaced parallel lines

AB = a and BC = b
1. Express
1. AC in terms of a and b
2. AD in terms of a and b
2. Using triangle BEP, express BP in terms of a and b
3. PR produced meets BA produced at X and PR = 1/9b – 8/3a
By writing PX as kPR and BX as hBA and using the triangle BPX determine the ratio PR: RX
20. The position vectors of points x and y are x = 2i + j – 3k and y = 3i + 2j – 2k respectively. Find XY 2. Given that X = 2i + j− 2K, y = -3i + 4j – k and z= 5i + 3j + 2k and that p= 3x – y + 2z, find the magnitude of vector p to 3 significant figures.

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