- Vectors in 3 Dimensions
- Column and Position Vectors
- Magnitude of a 3 Dimensional Vector.
- Distance Formula for a 3 Dimensional Vector
- Parallel Vectors and Collinearity
- Proportional Division of a Line
- The Ratio Theorem
- Past KCSE Questions on the Topic

## 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| = √(x
^{2}+y^{2 }+z^{2}) - This is the length of the vector.
- Use Pythagoras’ Theorem in 3 dimensions.

AB^{2 }= AR^{2 }+ BR^{2}= (AP^{2 }+ PR^{2}) + BR^{2}

=(x_{B}− x_{A})^{2}+ (y_{B}− y_{A})^{2}+ (z_{B}− z_{A})^{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| = √(x
^{2}+y^{2 }+z^{2}) - Since x = x
_{B }−_{}x_{A}and y = y_{B }−_{}y_{A}and z_{B }− z_{A}

**Example:**

- If A is (1, 3, 2) and B is (5, 6, 4)

→

Find |AB| - 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

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

**Solution**

**AB = OA + OB**

= −**a**+**b**

AC = −**a**+ 3**b**- OM =OA + AM

= OA + ¾AB

=**a**+¾(−**a**+**b**)

=**a**− ¾**a**+¾**b**

= ¼**a**+ ¾**b**

ON =OA +AN

=OA + ½AC**a**+ ½(−**a**+ 3**b**)**a**=**a**− ½**a**+^{3}/_{2}**b**

= ½**a**+^{3}/_{2}**b** - OM = kON ¾
**b**+ ¼**a**=^{k}/_{2}**a**+^{3k}/_{2}**b**

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 **– 3**a **= 4**b **– 4**p**

7**p** = 4**b **+ 3**a**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 = m AB

m+n

Therefore, OS = OA + m AB

m+n

Thus S = **a** + m (−**a** + **b**)

m+n

= **a** − m **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 **a **+ 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** + 3 **b**

5+3 5+3

Thus, r = ^{3}/_{8}a + ^{3}/_{8}b

**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 **q ** + 7 **r**

7+(−3) 7+(−3)

P = −^{3}/_{4} **q** + ^{7}/_{4}**r**

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:

- OM : MQ
- 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}/_{4}**a**) + ( 1−n )**b**

Comparing the co-efficients^{5}/_{7} k = ^{3}/_{4}n………..1^{2}/_{7}k = 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

- 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**- Express
- AV in terms of
**a**and**c** - BV in terms of
**a**,**b**and**c**

- AV in terms of
- M is point on OV such that OM: MV=3:4, Express BM in terms of
**a**,**b**and**c**. Simplify your answer as far as possible

- Express
- In triangle OAB, OA =
**a,**OB =**b**and P lies on AB such that AP: BP = 3.5- Find the terms of
**a**and**b**the vectors- AB
- AP
- BP
- OP

- Point Q is on OP such AQ = −
^{5}/_{8}**a**+^{9}/_{40}**b**Find the ratio OQ: QP

- Find the terms of
- 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
- Given that OA =
**a**and OB =**b**, express in terms of a and b:- AN
- BM

- If AX =
**s**AN and BX =**t**BM, 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**

- Given that OA =
- 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.
- In the figure below, vector OP =
**p**and OR =**r**. Vector OS = 2**r**and OQ =^{3}/_{2}**p**.- Express in terms of p and r (i) QR and (ii) PS
- The lines QR and PS intersect at K such that QK =
**m**QR and PK =**n**PS, 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**. - State the ratio PK: KS

- 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
- 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.
- The points P, Q, R and S have position vectors 2
**p**, 3**p**,**r**and 3**r**respectively, relative to an origin O. A point T divides PS internally in the ratio 1:6- Find, in the simplest form, the vectors OT and QT in terms p and r
- Show that the points Q, T, and R lie on a straight line
- Determine the ratio in which T divides QR

- Two points P and Q have coordinates (-2, 3) and (1 , 3) respectively. A translation map point P to P’ (10, 10)
- Find the coordinates of Q’ the image of Q under the translation
- 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

- Given that
**q**i +^{1}/_{3}j +^{2}/_{3 }k is a unit vector, find**q** - 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
- Find vector LN
- Given that a point M is on LN such that LM: MN = 3: 4, find the coordinates of M
- If line OM is produced to T such that OM: MT = 6:1
- Find the position vector of T
- Show that points L, T and B are collinear

- 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.- Express in terms of q and r
- XR
- YQ

- If XE =
**m**XR and YE =**n**YQ, express OE in terms of:- r, q and m
- r, q and n

- Using the results in (b) above, find the values of m and n.

- Express in terms of q and r
- 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.
- 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 - 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 2**u**=**v**+**w** - 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
- Position vector of Q
- Distance of Q from the origin

- 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.
- In the figure below OA =
**a**, OB =**b**, AB = BC and OB: BD = 3:1- Determine
- AB in terms of
**a**and**b** - CD, in terms of
**a**and**b**

- AB in terms of
- If CD: DE = 1:k and OA: AE = 1:m determine
- DE in terms of
**a**,**b**and**k** - The values of
**k**and**m**

- DE in terms of

- Determine
- The figure below shows a grid of equally spaced parallel lines

AB =**a**and BC =**b**- Express
- AC in terms of
**a**and**b** - AD in terms of
**a**and**b**

- AC in terms of
- Using triangle BEP, express BP in terms of
**a**and**b** - PR produced meets BA produced at X and PR =
^{1}/_{9}b –^{8}/_{3}a

By writing PX as**k**PR and BX as**h**BA and using the triangle BPX determine the ratio PR: RX

- Express
- 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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