vectors

Vectors

Introduction• This chapter focuses on vectors

• Vectors are used to describe movement in a given direction

• They are also used to describe straight lines in 3D (in a similar way to y = mx + c being used for 2D straight line graphs)

Teachings for Exercise 5A

VectorsYou need to know the difference between a scalar and a vector, and how to write down vectors

and draw vector diagrams

A scalar quantity has only a magnitude (size)

A vector quantity has both a magnitude and a direction

5A

Scalar VectorThe distance from P to Q is

100m P

Q

From P to Q you go 100m

north

N

60°

ScalarA ship is sailing

at 12km/h

VectorA ship is sailing at 12km/h on a bearing of 060°Direction and

Magnitude

http://www.youtube.com/watch?v=KbrEBpCw3Ag

http://www.youtube.com/watch?v=KbrEBpCw3Ag



Equal vectors have the same magnitude and direction

5A

P

Q

R

S

PQ = RSA Common way of showing vectors is using the letters

with an arrow above

a

b

Alternatively, single letters can be used…



Two vectors can be added using the ‘Triangle Law’

5A

a b

a + b

It is important to note that vector a + b is the single line from the start of a to

the end of b.

Vector a + b is NOT the two separate lines!

VectorsYou need to know the difference

between a scalar and a vector, and how to write down vectors and

draw vector diagrams

Draw a diagram to show the vector a + b + c

5A

a b

a + b + c

c

a b

c



Adding the vectors PQ and QP gives a Vector result of 0.

Vectors of the same size but in opposite directions have opposite signs (eg) + or -

5A

P

Q

P

Q

a

-a



The modulus value of a vector is another name for its magnitude

Eg) The modulus of the Vector a is |a|

The modulus of the vector PQ is |PQ|

Question: The vector a is directed due east and |a| = 12. Vector b is directed due south and |b| = 5. Find |a + b|

5A

a

ba + b

|𝑎+𝑏|2=122+52

|𝑎+𝑏|2=169

|𝑎+𝑏|=13

Use Pythagoras’ Theorem

Square the shorter sides…Square

Root



In the diagram opposite, find the following vectors in terms of a, b, c and d.

a) PS

b) RP

c) PT

d) TS

5A

P

Q

R

T

S

abc

d

= -a+ c Or c - a

= -b+ a Or a - b

= -a+ b+ d Or b + d - a

= -d- b+ c Or c - b - d

Teachings for Exercise 5B

VectorsYou need to be able to perform simple vector

arithmetic, and know the definition of a unit vector

The diagram shows the vector a. Draw diagrams to show the

vectors 3a and -2a

Vector 3a will be in the same direction as a, but 3 times the size

Vector -2a will be twice as big as s, but also in the opposite

direction

5B

a

3a

-2a



Any vector parallel to a may be written as λa, where λ (lamda) is a non-zero scalar (ie - represents a

number…)

Show that the vectors 6a + 8b and 9a + 12b are parallel…

5B

6𝒂+8𝒃 9𝒂+12𝒃32

(6𝒂+8𝒃 )Factorise

The second Vector is a multiple of the first, so they are parallel.

In this case, λ is 3/2 or 1.5



A unit vector is a vector which has a magnitude of 1 unit

Vector a has a magnitude of 20 units. Write down a unit vector

that is parallel to a.

5B

The unit vector will be:120 𝒂

This will be in the same direction as a with a magnitude of 1 unit

As a general rule, divide any vector by its magnitude to obtain

a parallel unit vector

¿𝑎|𝑎|



If:

And the vectors a and b are not parallel and non-zero, then:

and

Effectively, if the two vectors are equal then the coefficients of a

and b must also be equal

5B

λ 𝒂+𝜇𝒃=𝛼 𝒂+𝛽𝒃

λ=𝛼 𝜇=𝛽

Given that:5𝒂−4𝒃=(2𝑠+𝑡 )𝒂+(𝑠−𝑡 )𝒃Find the values of the scalars s and t

Comparing coefficients:

2𝑠+𝑡=5𝑠−𝑡=−4

1)2)

3 𝑠=1𝑠=13

𝑡=4 13

Add the equations together

Divide by 3

Sub into either of 1) or 2) to find the value of t

VectorsYou need to be able to perform

simple vector arithmetic, and know the definition of a unit vector

In the diagram opposite, PQ = 3a, QR = b, SR = 4a and PX = kPR. Find in terms of a, b and k:

a) PS

b) PX

c) SQ

d) SX

5B

P Q

RS

X

3a

4a

bb-a

= 3a + b – 4a

= b – a

= kPR

= k(3a + b)

= 4a - b

= -b + a

k(3a+b)

+ k(3a + b)= -b + a + 3ka +

kb= (3k + 1)a + (k – 1)b

Multiply out the bracketGroup up and factorise the ‘a’ and ‘b’ terms

separately

VectorsYou need to be able to perform simple vector arithmetic, and know the definition of a unit

vector

5B

P Q

RS

X

3a

4a

b

e) Use the fact that X lies on SQ to find the value of kSQ = 4a - b SX = (3k + 1)a + (k – 1)b

Since X is on SQ, SX and SQ are parallel, ie) one is a multiple of another!

(3𝑘+1 )𝒂+(𝑘−1 )𝒃¿ λ (4𝒂−𝒃) Use the lamda symbol to represent

one being a multiple of the

other…(3𝑘+1 )𝒂+(𝑘−1 )𝒃¿ 4 λ𝒂− λ𝒃3𝑘+1=4 λ𝑘−1=− λ

3𝑘+1=4 λ4𝑘−4=−4 λ

1)2) x4

7𝑘−3=0𝑘=

37

Add together

Solve for k

Multiply out the bracket

Teachings for Exercise 5C

VectorsYou need to be able to use

vectors to describe the position of a point in 2 or 3 dimensions

The position vector of a point A is the vector OA, where O is the

origin. OA is often written as a.

AB = b – a, where a and b are the position vectors of A and B

respectively.

5C

O

A

a

aB

A

b

b - a

O

VectorsYou need to be able to use

vectors to describe the position of a point in 2 or 3 dimensions

In the diagram, points A and B have position vectors a and b

respectively. The point P divides AB in the ratio 1:2.

Find the position vector of P.

5C

A

B

P

O

a

b

1

2b - a

𝐴𝐵=𝒃−𝒂 Using the rule we just saw…

1/3(b – a)2/3(b – a)

If the line is split in the ratio 1:2, then one part is 1/3 and the other is 2/3

𝑂𝑃=𝒂+13 (𝒃−𝒂)

𝑂𝑃=23 𝒂+

13 𝒃 The position vector of P is how we get from O to P

Teachings for Exercise 5D

VectorsYou need to know how to write

down and use the Cartesian components of a vector in 2

dimensions

The vectors i and j are unit vectors parallel to the x and y axes, in the

increasing directions

The points A and B in the diagram have coordinates (3,4) and (11,2)

respectively. Find in terms of i and j:

a) OA

b) OB

c) AB

5D

5 10 1500

5

10

A

Ba

b¿3 𝒊+4 𝒋¿11𝒊+2 𝒋¿𝒃−𝒂¿ (11𝒊+2 𝒋 )−(3 𝒊+4 𝒋 )¿8 𝒊−2 𝒋



dimensions

You can write a vector with Cartesian components as a column matrix:

Column matrix notation can be easier to read and avoids the need to write

out lots of i and j terms.

5D

𝑥 𝒊+𝑦 𝒋=(𝑥𝑦 )

Given that:a = 2i + 5jb = 12i – 10jc = -3i + 9j

Find a + b + c

𝒂+𝒃+𝒄=¿(25)+( 12−10)+(−39 )𝒂+𝒃+𝒄=¿(114 )

Be careful

with negatives

!

¿11𝒊+4 𝒋



dimensions

The modulus (magnitude) of xi + yj is:

This comes from Pythagoras’ Theorem

5D

√𝑥2+ 𝑦2

xi

yjxi + yj

The vector a is equal to 5i - 12j. Find |a| and find a unit vector in the same direction as a.

5i – 12j

5i

12j

|𝒂|=√52+ (−12 )2

|𝒂|=√169|𝒂|=13

¿𝒂

|𝒂|𝑈𝑛𝑖𝑡 𝑣𝑒𝑐𝑡𝑜𝑟 ¿5 𝒊−12 𝒋13

¿113 (5 𝒊−12 𝒋 )

¿ 113 ( 5−12)

Alternative notation…



dimensions

The modulus (magnitude) of xi + yj is:

5D

√𝑥2+ 𝑦2

Given that a = 5i + j and b = -2i – 4j, find the exact value of |2a + b|

2𝒂+𝒃=¿2(51)+(−2−4)¿ (102 )+(−2−4)¿ ( 8−2)

¿2𝒂+𝒃∨¿√82+(−2)2

¿√6 8¿2√17

Use x = 8 and y = -

2

‘Exact’ means you can leave in surd form

Teachings for Exercise 5E

VectorsYou need to know how to use

Cartesian coordinates in 3 dimensions

Cartesian coordinates in three dimensions are usually referred to as

the x, y and z axes, each at right-angles to the other.

Coordinates in 3 dimensions are given in the form (x, y, z)

5Ex

y

z

Imagine the x and y-axes have fallen down

flat, and the z-axis sticks up vertically out

of the origin…

Find the distance from the origin to the point P(4, 2, 5)

y

z

x4

5

2

You can use the 3D version of Pythagoras’ Theorem

The distance from the origin to the point (x, y, z) is given by:√𝑥2+𝑦2+𝑧 2

¿√ 42+22+52¿6.71(2dp)






5Ex

y

z

Find the distance between the points A(1, 3, 4) and B(8, 6, -5)

𝐴𝐵=¿𝒃−𝒂¿ ( 86−5)−(134)¿ ( 73−9)

¿ 𝐴𝐵∨¿√72+32+(−9)2

First calculate the vector from A to B

Then use 3D Pythagoras

¿√139¿11.8(1dp)






5Ex

y

z

The coordinates of A and B are (5, 0, 3) and (4, 2, k) respectively. Given that |AB| is 3 units, find the possible

values of k

𝐴𝐵=(42𝑘)−(503)𝐴𝐵=( −12𝑘−3)|𝐴𝐵|=√(−1)2+22+(𝑘−3)2

|𝐴𝐵|=√𝑘2−6𝑘+143=√𝑘2−6𝑘+149=𝑘2−6𝑘+140=𝑘2−6𝑘+50=(𝑘−5)(𝑘−1)𝑘=5 𝑜𝑟 𝑘=1

Calculate AB using k

Use Pythagoras in 3D

Careful when squaring the bracket

|AB| = 3

Square both sides

Solve as a quadratic

Teachings for Exercise 5F

VectorsYou can extend the two

dimensional vector results to 3 dimensions, using k as the unit

vector parallel to the z-axis

The vectors i, j and k are unit vectors parallel to the x, y and z-axes in the increasing directions

The vector xi + yj + zk can be written as a column matrix:

The modulus (magnitude) of xi + yj + zk is given by:

5F

(𝑥𝑦𝑧 )

√𝑥2+𝑦2+𝑧 2

The points A and B have position vectors 4i + 2j + 7k and 3i + 4j – k respectively. Find |AB| and show

that triangle OAB is isosceles.

¿ 𝐴𝐵∨¿ √(−1)2+22+(−8)2

𝐴𝐵=( 34−1)−(427 )𝐴𝐵=(−12−8)

𝐴𝐵=𝒃−𝒂

¿ 𝐴𝐵∨¿ √69

¿𝑂𝐴∨¿√ 42+22+72¿𝑂𝐴∨¿√69¿𝑂𝐵∨¿ √32+42+(−1)2

¿𝑂𝐵∨¿ √26

Find the vector AB

Now find the magnitude of

AB

Find the magnitude of OA and OB using

their position vectors

Isosceles as 2 vectors are equal…




The points A and B have the coordinates (t, 5, t-1) and (2t, t, 3)

respectively.

a) Find |AB|

b) By differentiating |AB|2, find the value of t for which |AB| is a minimum

c) Hence, find the minimum value of |AB|

5F

a) Find |AB|

𝐴𝐵=𝒃−𝒂𝐴𝐵=(2 𝑡𝑡3 )−( 𝑡

5𝑡−1)

𝐴𝐵=( 𝑡𝑡−54−𝑡)

|

|

|

Careful with the bracket

expansion!

Calculate the vector AB

Find the magnitude of AB in terms of t





respectively.

a) Find |AB|



5F


|

|

|

𝑑𝑝𝑑𝑡 =6 𝑡−18

0=6 𝑡−183=𝑡

Square both sides

Differentiate (often p is used to represent the

vector)Set equal to 0 for a

minimumSolve

It is possible to do this by differentiating |AB| rather than |AB|2, but it can be more difficult!

The value of t = 3 is the value for which the distance between the points A and B is the

smallest..





respectively.

a) Find |AB|



5F


|

𝑡=3

|

|

|

| (2dp)

Sub in the value of t

So for the given coordinates, the closest that points A and B could be is 3.74 units apart, when

t = 3.

Teachings for Exercise 5G

VectorsYou need to know the definition of

the scalar product of two vectors in 2 or 3 dimensions, and how it can be used to calculate the angle between

2 vectors

On the diagram to the right, the angle between a and b is θ. The two vectors must be directed away from point X

On the second diagram, vector b is directed towards X. Hence, the angle

between the two vectors is 160°.

This comes from re-drawing the diagram with vector b pointing away from point X.

5G

X

a

b

30°

X

a

b20°

X

a

b

20°160°

b


the scalar product of two vectors in 2 or 3 dimensions, and how it can

be used to calculate the angle between 2 vectors

The scalar product of two vectors a and b is written as a.b (‘a dot b’) and is

defined by:

The scalar product can be thought of as ‘the effect of one of the two vectors on

the other’

5G

𝒂 .𝒃=|𝒂|∨𝒃∨𝑐𝑜𝑠 𝜃

Vector multiplication

a.b = |a||b| By multiplication

a b

ab

In this case, the vector a can be split into a

horizontal and vertical component

Here we only consider the horizontal component

as this is in the direction of vector b

|a|cosθθ

By GCSE trigonometry

a.b = |a|cosθ|b|

a.b = |a||b|cosθ

This is the formula for the scalar ‘dot’

product of 2 vectors





defined by:

This formula can be rewritten in order to find the angle between 2 vectors:

5G


𝑐𝑜𝑠𝜃=𝒂 .𝒃

|𝒂|∨𝒃∨¿¿

ab

If two vectors are perpendicular, then the angle between them is 90°. As cos90° = 0, this will cause the dot

product to be 0 as well

Hence, if vectors are perpendicular, the dot product is 0

If the dot product is 0, the vectors are perpendicular

VectorsYou need to know the definition of the scalar product of two vectors in 2 or 3 dimensions, and how it can be used to calculate the angle between 2 vectors

The scalar product of two vectors a and b is written as a.b (‘a dot b’) and is defined by:


If we are to use this formula to work out the angle between 2 vectors, we therefore need

an alternative way to calculate the scalar product…

5G



|𝒂|∨𝒃∨¿¿

If a = x1i + y1j + z1kand b = x2i + y2j + z2k

Then:

𝒂 .𝒃=(𝑥1𝑦1𝑧1) .(𝑥2𝑦2𝑧2)¿ 𝑥1𝑥2+ 𝑦1 𝑦2+𝑧1 𝑧2

This is a way to find the dot product from 2 vectors





defined by:


5G



|𝒂|∨𝒃∨¿¿

If a = x1i + y1j + z1kand b = x2i + y2j + z2kThen:

𝒂 .𝒃=(𝑥1𝑦1𝑧1) .(𝑥2𝑦2𝑧2)¿ 𝑥1𝑥2+ 𝑦1 𝑦2+𝑧1 𝑧2

Given that a = 8i – 5j – 4k and b = 5i + 4j – k:a) Find a.b

𝒂 .𝒃=(𝑥1𝑦1𝑧1) .(𝑥2𝑦2𝑧 2)

𝒂 .𝒃=( 8−5−4 ) .(54−1)

𝒂 .𝒃=(8×5 )+ (−5×4 )+(−4×−1)𝒂 .𝒃=24

Use the dot product formula





defined by:


5G



|𝒂|∨𝒃∨¿¿


𝒂 .𝒃=(𝑥1𝑦1𝑧1) .(𝑥2𝑦2𝑧2)¿ 𝑥1𝑥2+ 𝑦1 𝑦2+𝑧1 𝑧2

Given that a = 8i – 5j – 4k and b = 5i + 4j – k:a) Find a.b 𝒂 .𝒃=24

b) Calculate the angle between vectors a and b


|𝒂|∨𝒃∨¿¿

|a |b

|a |b


|𝒂|∨𝒃∨¿¿

𝑐𝑜𝑠𝜃=24

√105√42

𝜃=68.8 °

Use the angle formula – you will need to calculate the

magnitude of each vector as well…

Sub in the values

Solve, remembering to use inverse Cos





defined by:


5G



|𝒂|∨𝒃∨¿¿


𝒂 .𝒃=(𝑥1𝑦1𝑧1) .(𝑥2𝑦2𝑧2)¿ 𝑥1𝑥2+ 𝑦1 𝑦2+𝑧1 𝑧2

Given that the vectors a = 2i – 6j + k and b = 5i + 2j + λk are perpendicular, calculate the value of λ.

𝒂 .𝒃=(𝑥1𝑦1𝑧1) .(𝑥2𝑦2𝑧 2)

𝒂 .𝒃=( 2−61 ) .(52λ)𝒂 .𝒃=(2×5 )+(−6×2 )+(1×λ)𝒂 .𝒃=−2+ λ0=−2+λλ=2

Calculate the dot product in terms of λ

As the vectors are perpendicular, the dot product

must be 0Solve

Only this value of λ will cause these vectors to be perpendicular…




5G

Given that a = -2i + 5j - 4k and b = 4i - 8j + 5k, find a vector which is

perpendicular to both a and b

Let the required vector be xi + yj + zk

𝒂 .(𝑥𝑦𝑧)=0 𝒃 . (𝑥𝑦𝑧 )=0andThe dot products of

both a and b with the required vector will be

0

(−25−4 ). (𝑥𝑦𝑧 )=0 ( 4−85 ) .(𝑥𝑦𝑧 )=0

−2 𝑥+5 𝑦−4 𝑧=0 4 𝑥−8 𝑦+5 𝑧=0−2 𝑥+5 𝑦=4 4 𝑥−8 𝑦=−5

Let z = 1

Let z = 1

−4 𝑥+10 𝑦=84 𝑥−8 𝑦=−5

2 𝑦=3𝑦=

32 𝑥=

74

Now solve as simultaneous

equations

So a possible answer would be:

x2

74 𝒊+

32 𝒋+𝒌 7 𝒊+6 𝒋+4𝒌

x4

Choosing a different value for z will lead to a vector that is a different size, but which is still pointing in the same direction

(ie – perpendicular)

However, this will not work if you choose z = 0




5G

Given that a = -2i + 5j - 4k and b = 4i - 8j + 5k, find a vector which is

perpendicular to both a and b7 𝒊+6 𝒋+4𝒌

The 3D axes show the 3 vectors in question. The green vector is perpendicular to both the others,

but you can only see this clearly when it is rotated!




5G

On this example, the second picture shows the diagram being viewed from the top of the

red vector




5G

On this example, the second picture shows the diagram being viewed from the top of the

red vector The vectors do not need to be touching – it

is always possible to find a vector that is perpendicular to 2 others!

Teachings for Exercise 5H

VectorsYou need to be able to write

the equation of a straight line in vector form (effectively the

equation of a 3D line!)

Let us first consider how this is done in 2 dimensions

So any linear 2D graph needs a direction, and a point on the line

With just the direction, the line wouldn’t have a specific path and

could effectively be anywhere

With only a given point, the line would not have a specific direction

5H

y

x

𝑦=𝑚𝑥+𝑐

m is the gradient of the line This can also be thought of as

the DIRECTION the line goes

c is the y-intercept This is a given point

on the line

VectorsYou need to be able to write the equation of a straight

line in vector form (effectively the equation of a

3D line!)

In 3D, we effectively need the same bits of information

We need any point on the line (ie – a coordinate in the form

(x, y, z))

We also need to know the direction the line is travelling (a vector with terms i, j and

k)

5H

A vector equation of a straight line passing through the point A with position vector a (effectively the

coordinate), and parallel to the vector b, is:

𝒓=𝒂+𝑡𝒃where t is a scalar

parameter



3D line!)

5H

A vector equation of a straight line passing through the point A with position vector a (effectively the coordinate), and parallel to the

vector b, is:


parameter

Find a vector equation of the straight line which passes through a, with position vector 3i – 5j + 4k,

and is parallel to the vector 7i – 3k

𝒂=( 3−54 ) 𝒃=( 70−3)This is the position vector we will use

This is the direction vector we will use

𝒓=𝒂+𝑡𝒃

𝒓=¿( 3−54 )+𝑡 ( 70−3) This is the vector equation of the

lineThe value t remains unspecified at this point, it can be used later to calculate points on the vector itself,

by substituting in different values for t



3D line!)

5H

A vector equation of a straight line passing through the point A with position vector a (effectively the coordinate), and parallel to the

vector b, is:


parameter

𝒓=¿( 3−54 )+𝑡 ( 70−3)Some alternative

forms𝒓=3 𝒊−5 𝒋+4𝒌+𝑡 (7 𝒊−3𝒌)

𝒓=(3 𝒊+7 𝑡 ) 𝒊+(−5) 𝒋+(4−3 𝑡 )𝒌

𝒓=( 3+7 𝑡−54−3 𝑡)

(By writing in a different form)

(By multiplying out the brackets and then re-grouping i, j and k terms)

(By rewriting again in the original column vector form)



3D line!)

5H

A vector equation of a straight line passing through the points A and B, with position vectors a and b respectively, is

given by:𝒓=𝒂+𝑡 (𝒃−𝒂)where t is a scalar parameter

As you aren’t given the direction vector in this type, you have to work it out by

calculating the vector AB (b – a)

Find a vector equation of the straight line passing through the points A and B, with coordinates (4, 5, -1)

and (6, 3, 2) respectively.

𝒂=( 45−1) 𝒃=(632)𝒃−𝒂=(632)−( 45−1)𝒃−𝒂=( 2−23 )

Calculating b – a will give you the vector AB, ie) the direction vector that passes through A

and B

𝒓=𝒂+𝑡 (𝒃−𝒂)

𝒓=( 45−1)+𝑡(2−23 )

Then use (b – a) along with either of the 2 coordinates/position vectors you’re given

Working in 2D – the equation of the line can be calculated by using either:a) The gradient (direction) and a

coordinate (like we just did)b) Two coordinates (since you can

calculate the gradient between them)

3D can also be done either way…



3D line!)

5H





The straight line l has a vector equation: r = (3i + 2j – 5k) + t(i – 6j – 2k)

Given that the point (a, b, 0) lies on l, calculate the values of a and b

𝑟=( 32−5)+𝑡 (1−6−2)

3+𝑡=𝑎2−6 𝑡=𝑏−5−2𝑡=0

The top numbers give the x coordinate, the

middles give the y, and the bottom gives the z,

all for an unknown value of t (at this point)

We can use the bottom equation to find the value of t

−5−2𝑡=0𝑡=−2.5

3+𝑡=𝑎 2−6 𝑡=𝑏3+(−2.5)=𝑎0.5=𝑎

2−6 (−2.5)=𝑏17=𝑏

The coordinate itself is (0.5, 17, 0)



3D line!)

5H





The straight line l has vector equation:r = (2i + 5j – 3k) + t(6i – 2j + 4k)

Show that an alternative vector equation of l is:r = (8i + 3j + k) + t(3i – j + 2k)

𝑟=( 25−3)+𝑡 (6−24 ) 𝑟=(831)+𝑡 (

3−12 )

If you look at the direction vectors, one is a multiple of the other

This means they are parallel and hence it does not matter which you use…

𝑟=( 25−3)+𝑡 (3−12 )

Original vector

updated with a different ‘b’

part

𝑟=( 25−3)+( 6−24 )If t = 2

𝑟=(831) So a coordinate on the line is (8, 3, 1)

1) Rewrite the original straight line equation with a different direction

vector2) Then try to find a value for t that will

give you the given coordinate as an answer

This shows that the given coordinate is on the line and hence, can be used in

the vector equation𝑟=(831)+𝑡 (

3−12 )

Teachings for Exercise 5I

VectorsYou need to be able to determine whether two given straight lines

intersect

Up until now we have used t as the scalar parameter

If we have more than one vector equation, then s is usually used to the other

Eg)

Sometimes the Greek letters λ and μ are used as well.

It is important to note that in 3 dimensions, 2 straight lines may pass each other without

intersecting!5I

𝒓=(5 𝒊+2 𝒋−3𝒌 )+𝑡 (2 𝒊−3 𝒋+𝒌)𝒓=(4 𝒊−5 𝒋+2𝒌 )+𝑠 (𝒊− 𝒋+6𝒌)

𝒓=(2 𝒊− 𝒋+2𝒌 )+λ (𝟒 𝒊−2 𝒋−2𝒌)

𝒓=(3 𝒊−5 𝒋+4𝒌 )+𝜇 (3 𝒊−3 𝒋+2𝒌)


intersect

It is important to note that in 3 dimensions, 2 straight lines may pass

each other without intersecting!

5I

1a) Show that the lines with vector equations:

r = (3i + 8j – 2k) + t(2i – j + 3k)

and r = (7i + 4j + 3k) + s(2i + j + 4k)

intersect.

𝑟=( 38−2)+𝑡 (2−13 )

𝑟=(743 )+𝑠(214)

3+2𝑡8− 𝑡−2+3 𝑡

7+2𝑠4+𝑠3+4 𝑠

Find the x, y and z coordinates in terms

of t and s

If there is a point of intersection, then at this point the equations for the x, y and z coordinates in terms of t and s

will be equal… Solve 2 of the equations simultaneously, and then check if

the answers also satisfy the third

3+2𝑡=7+2𝑠8− 𝑡=4+𝑠

2 𝑡−2 𝑠=4−𝑡−𝑠=−4𝑠=1 𝑡=3

Solve simultaneously by making either the t or s terms

‘equal’rearrange

−2+3 𝑡=3+4 𝑠−2+3(3)=3+4 (1)7=7

Sub s and t into the 3rd pair – if it ‘works’ then the lines intersect. If not, then they

don’t…

So the lines DO intersect


intersect

It is important to note that in 3 dimensions, 2 straight lines may pass

each other without intersecting!

5I

1a) Show that the lines with vector equations:

r = (3i + 8j – 2k) + t(2i – j + 3k)

and r = (7i + 4j + 3k) + s(2i + j + 4k)

intersect.

We have just calculated that the above lines intersect for the values of t = 3 and s = 1

b) Calculate the position vector of the point of intersection

𝑟=( 38−2)+𝑡 (2−13 )

𝑟=(743 )+𝑠(214)

𝑟=( 38−2)+3 (2−13 )

𝑟=(957)

𝑟=(743 )+1(214)

𝑟=(957)

Sub t = 3 into the first equation and

calculate the position vector

Sub s = 1 into the second equation and calculate the position

vector

You only need to choose one of the equations for the substitution, as you can see, it works for both!

Teachings for Exercise 5J

VectorsYou need to be able to calculate the angle between any 2 straight lines

The acute angle θ between two straight lines is given by:

Where a and b are the direction vectors of the two lines.

The lines do not have to be intersecting – the angle is the angle between them if

one was moved along so they do intersect

Eg) The lines to the right do not intersect, but the angle calculated is the angle

between them if one was translated such that they do intersect

5J

𝑐𝑜𝑠𝜃=¿

VectorsYou need to be able to calculate the angle between any 2 straight lines



The lines do not have to be intersecting – the angle is the angle

between them if one was moved along so they do intersect

5J

𝑐𝑜𝑠𝜃=¿

Modulus is used so that you get the acute angle rather than the obtuse

one

90 180

270 360

y = Cosθ

10

-1

For example, calculating cos-1(-0.5) would give us the angle 120°

For example, calculating cos-1|(-0.5)| would give us the angle 60° since -0.5 would be

replaced with 0.5 Each pair will always add up to 180°

Acute

Obtuse

This is because when 2 lines cross, you will always get a straight line with an acute and an obtuse angle on it

VectorsYou need to be able to calculate

the angle between any 2 straight lines



The lines do not have to be intersecting – the angle is the

angle between them if one was moved along so they do

intersect

5J

𝑐𝑜𝑠𝜃=¿

Find the acute angle between the lines with vector equations:

r = (2i + j + k) + t(3i – 8j – k)and r = (7i + 4j + k) + s(2i + 2j + 3k)

To do this, you only need the direction vectors

𝒂=( 3−8−1)𝒃=(223)𝒂 .𝒃=( 3−8−1) .(

223)

𝒂 .𝒃=(3×2 )+ (−8×2 )+(−1×3)𝒂 .𝒃=−13

¿𝒂∨¿√32+(−8)2+(−1)2

¿𝒂∨¿√74¿𝒃∨¿√22+22+32¿𝒃∨¿√17

Calculate the dot product, a.b

Calculate the magnitude of a and

b

VectorsYou need to be able to calculate

the angle between any 2 straight lines



The lines do not have to be intersecting – the angle is the

angle between them if one was moved along so they do

intersect

5J

𝑐𝑜𝑠𝜃=¿

𝒂 .𝒃=−13¿𝒂∨¿√74¿𝒃∨¿√17

𝑐𝑜𝑠𝜃=¿𝑐𝑜𝑠𝜃=| −13

√74 √17|𝑐𝑜𝑠𝜃=|−0.3665…|

𝑐𝑜𝑠𝜃=0.3665

𝜃=68.5 °

Sub in the values we have just calculated

Work out the sum

Since the answer is negative, we need to ‘make it positive’ by

multiplying by -1

Summary• We have learnt a great deal about vectors this chapter

• We have seen that when vectors are perpendicular, their dot ‘scalar’ product is equal to 0

• We have looked at the vector equation of a straight line

• We have seen how to calculate the angle between 2 lines

• We have learnt how to calculate whether 2 vectors intersect

vectors

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size vector

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vector quantity

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