Vectors Introduction This chapter focuses on vectors Vectors

Vectors Introduction  This chapter focuses on vectors  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 Vectors You need to know the difference between a scalar and a vector, and how to write down vectors and draw vector diagrams Q Scalar Vector The distance from P to Q is 100m From P to Q you go 100m north P A scalar quantity has only a magnitude (size) A vector quantity has both a magnitude and a direction

Scalar A ship is sailing at 12km/h Direction and Magnitude N 60 Vector A ship is sailing at 12km/h on a bearing of 060 5A Vectors Q You need to know the difference between a scalar and a vector, and how to write down vectors and draw vector diagrams S P Equal vectors have the same magnitude and direction A Common way of showing vectors is using the letters with an arrow above Alternatively, single letters can be used

R PQ = RS a b 5A Vectors You need to know the difference between a scalar and a vector, and how to write down vectors and draw vector diagrams Two vectors can be added using the Triangle Law b a 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! 5A Vectors You need to know the difference between a scalar and a vector, and how to write down vectors and draw vector diagrams a b

c Draw a diagram to show the vector a + b+c a a+b+c b c 5A Vectors Q You need to know the difference between a scalar and a vector, and how to write down vectors and draw vector diagrams a P Q 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 - -a P

5A Vectors You need to know the difference between a scalar and a vector, and how to write down vectors and draw vector diagrams The modulus value of a vector is another name for its magnitude a a+b b Use Pythagoras Theorem 2 Eg) The modulus of the Vector a is |a| | +| =12 2+5 2 The modulus of the vector PQ is |PQ| | +| =169 Question: The vector a is directed due east and |a| = 12. Vector b is directed due south and |b| = 5. Find | a + b| |+|=13 2 Square the shorter

sides Square Root 5A Vectors Q You need to know the difference between a scalar and a vector, and how to write down vectors and draw vector diagrams In the diagram opposite, find the following vectors in terms of a, b, c and d. a) PS a c P b S R d T = -a+ c Or c - a b) RP = -b+ a Or a - b c) PT = -a+ b+ d Or b + d - a d) TS

= -d- b+ c Or c - b - d 5A Teachings for Exercise 5B Vectors You 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 a 3a -2a Vector -2a will be twice as big as s, but also in the opposite direction 5B Vectors You need to be able to perform simple vector arithmetic, and know the definition of a unit vector Any vector parallel to a may be written as a, where (lamda) is a non-zero scalar (ie - represents a number)

Show that the vectors 6a + 8b and 9a + 12b are parallel 6 +8 9 +12 3 ( 6 + 8 ) 2 Factorise The second Vector is a multiple of the first, so they are parallel. In this case, is 3/2 or 1.5 5B Vectors You need to be able to perform simple vector arithmetic, and know the definition of a unit vector 1 The unit vector will be: 20 This will be in the same direction as a with a magnitude of 1 unit 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. As a general rule, divide any

vector by its magnitude to obtain a parallel unit vector || 5B Vectors Given that: You need to be able to perform simple vector arithmetic, and know the definition of a unit vector 5 4 =( 2 + ) +( ) Find the values of the scalars s and t If: + = + Comparing coefficients: 2 +=5 2) = 4 1) 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 3 =1 1 = 3 1 =4 3 Add the equations together Divide by 3 Sub into either of 1) or 2) to find the value of t 5B Vectors You need to be able to perform simple vector arithmetic, and know the definition of a unit vector In the diagram opposite, PQ = 3a, QR = P b-a 3a k(3a+b ) S

Q X 4a b R b, SR = 4a and PX = kPR. Find in terms of a, b and k: a) PS b) PX = 3a + b = b a 4a = k(3a + = b) kPR = 4a b = -b + + k(3a + d) SX a b) = -b + a + 3ka + kb = (3k + 1)a + (k 1)b c) SQ Multiply out the bracket Group up and factorise the a and b terms separately 5B Vectors

P You need to be able to perform simple vector arithmetic, and know the definition of a unit vector Q b X S e) Use the fact that X lies on SQ to find the value of k SQ = 4a - b 3a 4a R SX = (3k + 1)a + (k 1)b Since X is on SQ, SX and SQ are parallel, ie) one is a multiple of another! ( 3 +1 ) + ( 1 ) (4 ) ( 3 +1 ) + ( 1 ) 4 3 +1=4 2) 1= 1) x4 3 +1=4 4 4= 4 7 3=0

= 3 7 Multiply out the bracket Use the lamda symbol to represent one being a multiple of the other Add together Solve for k 5B Teachings for Exercise 5C Vectors A You need to be able to use vectors to describe the position of a point in 2 or 3 dimensions a The position vector of a point A is O the vector OA, where O is the origin. OA is often written as a. A a

AB = b a, where a and b are the B position vectors of A and B respectively. b-a O b 5C Vectors /3(b a) 1 P b-a 1 You 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 A 2 a B position vectors a and b b

respectively. The point P divides AB in the ratio 1:2. /3(b a) 2 O Find the position vector of P. = Using the rule we just saw If the line is split in the ratio 1:2, then one part is 1/3 and the other is 2/3 = + = 1 ( ) 3 2 1 + 3 3 The position vector of P is how we get from O to P 5C Teachings for Exercise 5D Vectors You 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 3 +4 1 0 5 A a B 0 b 0 5 10 15 11+2 ( 11+2 ) (3 +4 ) 8 2

5D Vectors You need to know how to write down and use the Cartesian components of a vector in 2 dimensions Given that: a = 2i + 5j b = 12i 10j c = -3i + 9j You can write a vector with Cartesian components as a column matrix: Find a + b + c + = ( ) Column matrix notation can be easier to read and avoids the need to write out lots of i and j terms. 1 2 + 3 + ++= ( 2 5 ) ( 10 ) ( 9 ) ++= ( 141) Be

careful with negatives ! 11+4 5D Vectors You need to know how to write down and use the Cartesian components of a vector in 2 dimensions The vector a is equal to 5i - 12j. Find |a| and find a unit vector in the same direction as a. 5i ||= 52 + ( 12 ) The modulus (magnitude) of xi + yj is: 2 + 2 5i 12j 12j 2 ||= 169 ||=13 xi + yj

yj xi This comes from Pythagoras Theorem 5 12 13 || 1 (5 12 ) 13 1 5 13 12 ( ) Alternative notation 5D Vectors You need to know how to write down and use the Cartesian components of a vector in 2 dimensions The modulus (magnitude) of xi + yj is: 2 +

2 Given that a = 5i + j and b = -2i 4j, find the exact value of |2a + b| 2+= 2 5 + 2 4 1 () ( ) 10 2 ( + 2 ) ( 4) ( 8 ) 2 2 + 8 2+( 2)2 6 8 2 17 Use x = 8 and y = 2 Exact means you can leave in surd form 5D Teachings for Exercise 5E Vectors You need to know how to use

Cartesian coordinates in 3 dimensions Find the distance from the origin to the point P(4, 2, 5) z y Cartesian coordinates in three dimensions are usually referred to as the x, y and z axes, each at rightangles to the other. Coordinates in 3 dimensions are given in the form (x, y, z) z 5 4 2 x y Imagine the x and yaxes have fallen down flat, and the z-axis sticks up vertically out of the origin x You can use the 3D version of Pythagoras Theorem The distance from the origin to the point (x, y, z) is2 given 2 by: 2 + + 42 +22 +52 6.71 (2dp)

5E Vectors You 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 rightangles to the other. Coordinates in 3 dimensions are given in the form (x, y, z) z y Find the distance between the points A(1, 3, 4) and B(8, 6, -5) = First calculate the vector from A to B 8 1 6 3 5 4 ( )() ( ) 7 3 9 72 +3 2 +( 9)2 Then use 3D Pythagoras

139 11.8 (1dp) x 5E Vectors You 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 rightangles to the other. Coordinates in 3 dimensions are given in the form (x, y, z) z y 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 4 5 = 2 0 3 () () ( ) 1 = 2 3 Calculate AB using k Use Pythagoras

in 3D | |=( 1)2 +2 2+( 3)2 Careful when squaring the bracket | |= 2 6 +14 |AB| = 3 2 3= 6 +14 2 9= 6 +14 0=2 6 +5 Square both sides Solve as a quadratic 0=(5)(1) x =5 =1 5E Teachings for Exercise 5F Vectors You 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 zaxes in the increasing directions The vector xi + yj + zk can be written as a column matrix:

() + 2 = Find the vector AB 3 4 4 2 1 7 ( )() ( ) = 1 = 2 8 (1)2 +22 +(8)2 69 The modulus (magnitude) of xi + yj + zk is given by: 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. + 2 4 2 +22 +7 2 69 2 2 2 3 +4 +(1) 26 Now find the magnitude of AB Find the magnitude of OA and OB using their position vectors Isosceles as 2 vectors are 5F Vectors You can extend the two dimensional vector results to 3 dimensions, using k as the unit vector parallel to the z-axis The points A and B have the coordinates (t, 5, t-1) and (2t, t, 3) respectively.

a) Find |AB| a) Find |AB| = 2 = 5 3 1 ( )( ) ( ) = 5 4 | b) By differentiating |AB|2, find the value of t for which |AB| is a minimum c) Hence, find the minimum value of |AB| Calculate the vector AB | Find the magnitude of AB in terms of t Careful with the bracket expansion!

| 5F Vectors You can extend the two dimensional vector results to 3 dimensions, using k as the unit vector parallel to the z-axis The points A and B have the coordinates (t, 5, t-1) and (2t, t, 3) respectively. | b) By differentiating |AB|2, find the value of t for which |AB| is a minimum | | a) Find |AB| =6 18 b) By differentiating |AB|2, find the value of t for which |AB| is a minimum 0=6 18 3= c) Hence, find the minimum value of |AB| Solve Square both sides

Differentiate (often p is used to represent the vector) Set equal to 0 for a minimum The value of t = 3 is the value for which the distance between the points A and B is the smallest.. It is possible to do this by differentiating |AB| rather than |AB|2, but it can be more difficult! 5F Vectors You can extend the two dimensional vector results to 3 dimensions, using k as the unit vector parallel to the z-axis | =3 c) Hence, find the minimum value of |AB| 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| Sub in the value of t | |

| | (2dp) So for the given coordinates, the closest that points A and B could be is 3.74 units apart, when t = 3. 5F Teachings for Exercise 5G Vectors a You 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 30 X b On the diagram to the right, the angle between a and b is . The two vectors must be directed away from point X a 20 X On the second diagram, vector b is directed towards X. Hence, the angle between the two vectors is 160. b

a 160 This comes from re-drawing the diagram with vector b pointing away from point X. b X 20 b 5G Vectors You 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 Vector multiplication a The scalar product of two vectors a and b is written as a.b (a dot b) and is defined by: b a.b = |a||b| By multiplication .=|| The scalar product can be thought of as

the effect of one of the two vectors on the other 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 b |a|cos a.b = |a|cos|b| By GCSE trigonometry a.b = |a||b|cos This is the formula for the scalar dot product of 2 vectors 5G Vectors You 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:

.=|| This formula can be rewritten in order to find the angle between 2 vectors: = . || a b 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 5G Vectors You 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 a = x1i + y1j + z1k and b = x2i + y2j + z2k Then: 1 2

. = 1 . 2 1 2 ( ) ( ) + + 1 2 1 2 1 2 .=|| This formula can be rewritten in order to find the angle between 2 vectors: = . || This is a way to find the dot product from 2 vectors 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 Vectors You 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 Given that a = 8i 5j 4k and b = 5i + 4j k: a) Find a.b 1 2 . = 1 . 2 1 2 ( )( ) ( )( ) The scalar product of two vectors a and b is written as a.b (a dot b) and is defined by: 8 5 . = 5 . 4 4 1 .=|| . = ( 8 5 ) + ( 5 4 ) +( 4 1) . =24 This formula can be rewritten in order to find the angle between . 2 vectors: = Use the dot product

formula || If a = x1i + y1j + z1k and b = x2i + y2j + z2k Then: 1 2 . = 1 . 2 1 2 ( )( ) + + 1 2 1 2 1 2 5G Vectors You 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 Given that a = 8i 5j 4k and b = 5i + 4j k: a) Find a.b b) Calculate the angle between vectors a and b

= The scalar product of two vectors a and b is written as a.b (a dot b) and is defined by: . || .=|| |a || = If a = x1i + y1j + z1k and b = x2i + y2j + z2k ( )( ) + + 1 2 1 2 1 2 |b |b . ||

= Then: 1 2 . = 1 . 2 1 2 Use the angle formula you will need to calculate the magnitude of each vector as well |a This formula can be rewritten in order to find the angle between . 2 vectors: = . =24 24 105 42 =68.8 Sub in the values Solve, remembering to use inverse Cos 5G Vectors Given that the vectors a = 2i 6j + k and b = 5i + 2j + k are perpendicular, calculate the value of .

You 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 1 2 . = 1 . 2 1 2 ( )( ) ( )( ) 2 5 . = 6 . 2 1 The scalar product of two vectors a and b is written as a.b (a dot b) and is defined by: .=|| . = ( 2 5 ) + ( 6 2 ) +(1 ) This formula can be rewritten in order to find the angle between . 2 vectors: = Calculate the dot product in terms of

|| If a = x1i + y1j + z1k and b = x2i + y2j + z2k . = 2+ 0=2+ =2 As the vectors are perpendicular, the dot product must be 0 Solve Then: 1 2 . = 1 . 2 1 2 ( )( ) + + 1 2 1 2 1 2 Only this value of will cause these vectors to be perpendicular 5G Vectors

You 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 Given that a = -2i + 5j - 4k and b = 4i - 8j + 5k, find a vector which is perpendicular to both a and b 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 Let the required vector be xi + yj + zk The dot products of . =0 and . = 0 both a and b with the required vector will be 0 2 4 . = 0 5 8 . =0 4

5 () () ( )( ) ( )( ) 2 +5 4 =0 2 +5 =4 Let z= 1 4 8 +5 =0 4 8 = 5 Let z= 1 x2 4 +10 =8 4 8 = 5 2 =3 = 3 2 Now solve as simultaneous equations =

7 4 So a possible answer would 7 3 be: 4 + 2 + x4 7+6 +4 5G Vectors Link to Autograph You 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 Given that a = -2i + 5j - 4k and b = 4i - 8j + 5k, find a vector which is perpendicular to both a and b 7+6 +4 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 Vectors Link to Autograph You 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 this example, the second picture shows the diagram being viewed from the top of the red vector 5G Vectors Link to Autograph You 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 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! 5G

Teachings for Exercise 5H Vectors You need to be able to write the equation of a straight line in vector form (effectively the equation of a 3D line!) y Let us first consider how this is done in 2 dimensions x So any linear 2D graph needs a direction, and a point on the line With just the direction, the line wouldnt have a specific path and could effectively be anywhere With only a given point, the line would not have a specific direction = + 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 5H Vectors You 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 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: =+ where t is a scalar parameter 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 Vectors You need to be able to write the equation of a straight line in vector form (effectively the equation of a 3D line!) 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: =+ where t is a scalar 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 3 = 5 4 ( ) This is the position vector we will use 7 = 0 3 ( ) This is the direction vector we will use =+ 3 7 5 + 0 4 3 ( ) ( ) = This is the vector equation of the line The 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 5H

Vectors You need to be able to write the equation of a straight line in vector form (effectively the equation of a 3D line!) 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: =+ where t is a scalar parameter 3 7 5 + 0 4 3 ( ) ( ) = Some alternative forms =3 5 +4 +(7 3 ) (By writing in a different form) =( 3 +7 ) +(5) +(4 3 ) (By multiplying out the brackets and then regrouping i, j and k terms) ( =

3 +7 5 43 ) (By rewriting again in the original column vector form) 5H Vectors You need to be able to write the equation of a straight line in vector form (effectively the equation of a 3D line!) 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 A vector equation of a straight line passing through the points A and B, with position vectors a and b respectively, is given by: =+( ) where t is a scalar parameter As you arent 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. 4 = 5 1 ( ) 6 = 3 2 () 6 4 = 3 5 1 2 ()( ) ( ) Calculating b a will give you the vector AB, ie) the direction vector that passes through A and B 2 = 2 3 =+( ) 4 2 = 5 + 2 1 3

( ) ( ) Then use (b a) along with either of the 2 coordinates/position vectors youre given 5H Vectors You need to be able to write the equation of a straight line in vector form (effectively the equation of a 3D line!) A vector equation of a straight line passing through the points A and B, with position vectors a and b respectively, is given by: =+( ) 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 3 1 = 2 + 6 5 2 ( ) ( ) 3+= 2 6 = 5 2=0

where t is a scalar parameter As you arent given the direction vector in this type, you have to work it out by calculating the vector AB (b a) 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 2=0 =2.5 3+= 2 6 = 3+(2.5)= 2 6 (2.5)= 0.5= 17= The coordinate itself is (0.5, 17, 0) 5H Vectors You need to be able to write the equation of a straight line in vector form (effectively the equation of a 3D line!)

A vector equation of a straight line passing through the points A and B, with position vectors a and b respectively, is given by: =+( ) where t is a scalar parameter As you arent given the direction vector in this type, you have to work it out by calculating the vector AB (b a) 1) Rewrite the original straight line equation with a different direction vector 2) 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 equation Original vector updated with a different b part 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) 2 6 + 5 2 3 4

( ) ( ) = 8 3 = 3 + 1 1 2 () ( ) 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 2 3 = 5 + 1 3 2 ( ) ( ) ( )( ) () () ( ) 2 6 = 5 + 2 3 4 8 = 3 1 8

3 = 3 + 1 1 2 If t = 2 So a coordinate on the line is (8, 3, 1) 5H Teachings for Exercise 5I Vectors Link to Autograph You 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) =( 5 +2 3 ) +(2 3 +) =( 4 5 +2 )+( +6 ) Sometimes the Greek letters and are used as well. =( 2 +2 )+ ( 2 2 ) =( 3 5 +4 ) +(3 3 +2 ) It is important to note that in 3 dimensions, 2 straight lines may pass each other without intersecting!

5I Vectors 3+2 8 ( ) ( ) 2+3 7+2 4+ ( ) ( ) 3+4 You need to be able to determine whether two given straight lines intersect It is important to note that in 3 dimensions, 2 straight lines may pass each other without intersecting! 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) 3 2 = 8 + 1 2 3 7 2 = 4 + 1 3 4 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+2=7 +2 8 =4 + rearrange intersect. 2+3 =3+4 2+3(3)=3+4 (1) 7=7 2 2 =4 = 4 =1 =3 Solve simultaneous ly by making either the t or s terms equal Sub s and t into the 3rd pair if it works then the lines intersect. If not, then they dont So the lines DO intersect 5I

Vectors ( ) ( ) ( ) ( ) () You need to be able to determine whether two given straight lines intersect It is important to note that in 3 dimensions, 2 straight lines may pass each other without intersecting! 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 3 2 = 8 + 1 2 3 = 3 2 +3 8 1 2 3 Sub t = 3 into the first

equation and calculate the position vector 9 = 5 7 7 2 = 4 + 1 3 4 () () () () () 7 2 = 4 +1 1 3 4 Sub s = 1 into the second equation and calculate the position vector 9 = 5 7 You only need to choose one of the equations for the substitution, as you can see, it works for both! 5I Teachings for Exercise 5J

Link to Autograph Vectors You need to be able to calculate the angle between any 2 straight lines The acute angle 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 Vectors You need to be able to calculate the angle between any 2 straight lines The acute angle 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 Modulus is used so that you get the acute angle rather than the obtuse one 1 y= Cos 0 -1 90 18 0 270 360 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 Obtuse Acute This is because when 2 lines cross, you will always get a straight line with an acute and an obtuse angle on it 5J Vectors You need to be able to calculate the angle between any 2 straight lines The acute angle 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 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 3 = 8 1 2 = 2 3 ( ) () ( )( ) 3 2 . = 8 . 2 1 3 Calculate the dot product, a.b . = ( 3 2 ) + ( 8 2 ) +(1 3)

.= 13 32 +(8) 2+(1)2 74 2 2+22 +32 17 Calculate the magnitude of a and b 5J Vectors You need to be able to calculate the angle between any 2 straight lines The acute angle between two straight lines is given by: = Where a and b are the direction vectors of the two lines. = | = 13 74 17 | =| 0.3665 |

=0.3665 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 =68.5 The lines do not have to be intersecting the angle is the angle between them if one was moved along so they do .= 13 intersect 74 17 5J 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

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