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

Direction and Magnitude 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 scalar quantity has only a magnitude (size)  A vector quantity has both a magnitude and a direction Scalar Vector The distance from P to Q is 100m From P to Q you go 100m north P N Scalar Vector A ship is sailing at 12km/h A ship is sailing at 12km/h on a bearing of 060° 60° Direction and Magnitude 5A

Alternatively, single letters can be used… Vectors Q You need to know the difference between a scalar and a vector, and how to write down vectors and draw vector diagrams  Equal vectors have the same magnitude and direction S P R A Common way of showing vectors is using the letters with an arrow above PQ = RS a Alternatively, single letters can be used… b 5A

Vector a + b is NOT the two separate lines! 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 Draw a diagram to show the vector a + b + c a b c a b c a + 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 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 Q -a P 5A

Vectors 𝑎+𝑏 2 = 12 2 + 5 2 𝑎+𝑏 2 =169 𝑎+𝑏 =13 5A a 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 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| b a + b Use Pythagoras’ Theorem 𝑎+𝑏 2 = 12 2 + 5 2 Square the shorter sides… 𝑎+𝑏 2 =169 Square Root 𝑎+𝑏 =13 5A

Vectors Q a 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. PS RP PT TS c b P S R d T = -a + c Or c - a = -b + a Or a - b = -a + b + d Or b + d - a = -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 Vector -2a will be twice as big as s, but also in the opposite direction a 3a -2a 5B

The second Vector is a multiple of the first, so they are parallel. 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 λa, where λ (lamda) is a non-zero scalar (ie - represents a number…) Show that the vectors 6a + 8b and 9a + 12b are parallel… 6𝒂+8𝒃 9𝒂+12𝒃 Factorise 3 2 6𝒂+8𝒃 The second Vector is a multiple of the first, so they are parallel.  In this case, λ is 3/2 or 1.5 5B

This will be in the same direction as a with a magnitude of 1 unit Vectors You need to be able to perform simple vector arithmetic, and know the definition of a unit vector 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. 1 20 𝒂 The unit vector will be: 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 5B

Vectors 5𝒂−4𝒃= 2𝑠+𝑡 𝒂+(𝑠−𝑡)𝒃 λ𝒂+𝜇𝒃=𝛼𝒂+𝛽𝒃 2𝑠+𝑡=5 𝑠−𝑡=−4 λ=𝛼 𝜇=𝛽 3𝑠=1 You need to be able to perform simple vector arithmetic, and know the definition of a unit vector 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 Given that: 5𝒂−4𝒃= 2𝑠+𝑡 𝒂+(𝑠−𝑡)𝒃 Find the values of the scalars s and t λ𝒂+𝜇𝒃=𝛼𝒂+𝛽𝒃 Comparing coefficients: 1) 2𝑠+𝑡=5 2) 𝑠−𝑡=−4 Add the equations together λ=𝛼 𝜇=𝛽 3𝑠=1 Divide by 3 𝑠= 1 3 Sub into either of 1) or 2) to find the value of t 𝑡=4 1 3 5B

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

Vectors 3𝑘+1 𝒂+ 𝑘−1 𝒃 = λ(4𝒂−𝒃) 3𝑘+1 𝒂+ 𝑘−1 𝒃 = 4λ𝒂−λ𝒃 3𝑘+1=4λ 3𝑘+1=4λ 3a P Q You need to be able to perform simple vector arithmetic, and know the definition of a unit vector b X R S 4a e) Use the fact that X lies on SQ to find the value of k SQ = 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! Use the lamda symbol to represent one being a multiple of the other… 3𝑘+1 𝒂+ 𝑘−1 𝒃 = λ(4𝒂−𝒃) Multiply out the bracket 3𝑘+1 𝒂+ 𝑘−1 𝒃 = 4λ𝒂−λ𝒃 1) 3𝑘+1=4λ 3𝑘+1=4λ x4 2) 𝑘−1=−λ 4𝑘−4=−4λ Add together 7𝑘−3=0 𝑘= 3 7 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 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. a O A b - a a B O b 5C

Vectors 𝐴𝐵=𝒃−𝒂 𝑂𝑃=𝒂+ 1 3 (𝒃−𝒂) 𝑂𝑃= 2 3 𝒂+ 1 3 𝒃 5C 1/3(b – a) 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 position vectors a and b respectively. The point P divides AB in the ratio 1:2. Find the position vector of P. 1 2/3(b – a) P b - a 2 a B b O 𝐴𝐵=𝒃−𝒂 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 𝑂𝑃=𝒂+ 1 3 (𝒃−𝒂) 𝑂𝑃= 2 3 𝒂+ 1 3 𝒃 The position vector of P is how we get from O to P 5C

Teachings for Exercise 5D

Vectors 5D =3𝒊+4𝒋 =11𝒊+2𝒋 =𝒃−𝒂 = 11𝒊+2𝒋 −(3𝒊+4𝒋) =8𝒊−2𝒋 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: OA OB AB 10 5 A a B b =3𝒊+4𝒋 5 10 15 =11𝒊+2𝒋 =𝒃−𝒂 = 11𝒊+2𝒋 −(3𝒊+4𝒋) =8𝒊−2𝒋 5D

Be careful with negatives! Vectors You need to know how to write down and use the Cartesian components of a vector in 2 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. Given that: a = 2i + 5j b = 12i – 10j c = -3i + 9j Find a + b + c 𝑥𝒊+𝑦𝒋= 𝑥 𝑦 2 5 + 12 −10 + −3 9 𝒂+𝒃+𝒄= Be careful with negatives! 11 4 𝒂+𝒃+𝒄= =11𝒊+4𝒋 5D

Alternative notation… 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: This comes from Pythagoras’ Theorem The vector a is equal to 5i - 12j. Find |a| and find a unit vector in the same direction as a. 5i 𝒂 = 5 2 + −12 2 𝑥 2 + 𝑦 2 12j 𝒂 = 169 5i – 12j xi + yj 𝒂 =13 yj = 𝒂 𝒂 = 5𝒊−12𝒋 13 𝑈𝑛𝑖𝑡 𝑣𝑒𝑐𝑡𝑜𝑟 xi = 1 13 (5𝒊−12𝒋) Alternative notation… = 1 13 5 −12 5D

‘Exact’ means you can leave in surd form 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: Given that a = 5i + j and b = -2i – 4j, find the exact value of |2a + b| 2 5 1 + −2 −4 2𝒂+𝒃= 𝑥 2 + 𝑦 2 = 10 2 + −2 −4 = 8 −2 Use x = 8 and y = -2 8 2 + (−2) 2 |2𝒂+𝒃|= = 68 ‘Exact’ means you can leave in surd form =2 17 5D

Teachings for Exercise 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 right-angles to the other. Coordinates in 3 dimensions are given in the form (x, y, z) Find the distance from the origin to the point P(4, 2, 5) z y 5 2 4 z x y You can use the 3D version of Pythagoras’ Theorem  The distance from the origin to the point (x, y, z) is given by: Imagine the x and y-axes have fallen down flat, and the z-axis sticks up vertically out of the origin… 𝑥 2 +𝑦 2 +𝑧 2 = 4 2 +2 2 +5 2 x =6.71 (2dp) 5E

Vectors 𝐴𝐵= 𝒃−𝒂 = 8 6 −5 − 1 3 4 = 7 3 −9 |𝐴𝐵|= 7 2 + 3 2 + (−9) 2 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 right-angles to the other. Coordinates in 3 dimensions are given in the form (x, y, z) Find the distance between the points A(1, 3, 4) and B(8, 6, -5) First calculate the vector from A to B 𝐴𝐵= 𝒃−𝒂 = 8 6 −5 − 1 3 4 = 7 3 −9 z y Then use 3D Pythagoras |𝐴𝐵|= 7 2 + 3 2 + (−9) 2 = 139 =11.8 (1dp) x 5E

Careful when squaring the bracket Vectors 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 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 right-angles to the other. Coordinates in 3 dimensions are given in the form (x, y, z) 𝐴𝐵= 4 2 𝑘 − 5 0 3 Calculate AB using k 𝐴𝐵= −1 2 𝑘−3 Use Pythagoras in 3D 𝐴𝐵 = (−1) 2 + 2 2 + (𝑘−3) 2 Careful when squaring the bracket z 𝐴𝐵 = 𝑘 2 −6𝑘+14 |AB| = 3 y 3= 𝑘 2 −6𝑘+14 Square both sides 9= 𝑘 2 −6𝑘+14 Solve as a quadratic 0= 𝑘 2 −6𝑘+5 0=(𝑘−5)(𝑘−1) x 𝑘=5 𝑜𝑟 𝑘=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 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: 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=𝒃−𝒂 Find the vector AB 𝐴𝐵= 3 4 −1 − 4 2 7 𝐴𝐵= −1 2 −8 𝑥 𝑦 𝑧 |𝐴𝐵|= (−1) 2 + 2 2 + (−8) 2 Now find the magnitude of AB |𝐴𝐵|= 69 |𝑂𝐴|= 4 2 + 2 2 + 7 2 Find the magnitude of OA and OB using their position vectors |𝑂𝐴|= 69 𝑥 2 + 𝑦 2 + 𝑧 2 |𝑂𝐵|= 3 2 + 4 2 + (−1) 2 |𝑂𝐵|= 26 Isosceles as 2 vectors are equal… 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. Find |AB| By differentiating |AB|2, find the value of t for which |AB| is a minimum Hence, find the minimum value of |AB| a) Find |AB| Calculate the vector AB 𝐴𝐵=𝒃−𝒂 𝐴𝐵= 2𝑡 𝑡 3 − 𝑡 5 𝑡−1 𝐴𝐵= 𝑡 𝑡−5 4−𝑡 Find the magnitude of AB in terms of t |𝐴𝐵|= 𝑡 2 + 𝑡−5 2 + 4−𝑡 2 Careful with the bracket expansion! |𝐴𝐵|= 𝑡 2 + 𝑡 2 −10𝑡+25+ 𝑡 2 −8𝑡+16 |𝐴𝐵|= 3𝑡 2 −18𝑡+41 5F

Vectors |𝐴𝐵|= 3𝑡 2 −18𝑡+41 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. Find |AB| By differentiating |AB|2, find the value of t for which |AB| is a minimum Hence, find the minimum value of |AB| b) By differentiating |AB|2, find the value of t for which |AB| is a minimum |𝐴𝐵|= 3𝑡 2 −18𝑡+41 Square both sides |𝐴𝐵 | 2 = 3𝑡 2 −18𝑡+41 Differentiate (often p is used to represent the vector) 𝑑𝑝 𝑑𝑡 =6𝑡−18 Set equal to 0 for a minimum 0=6𝑡−18 Solve 3=𝑡 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 |𝐴𝐵|= 3𝑡 2 −18𝑡+41 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. Find |AB| By differentiating |AB|2, find the value of t for which |AB| is a minimum Hence, find the minimum value of |AB| 𝑡=3 c) Hence, find the minimum value of |AB| |𝐴𝐵|= 3𝑡 2 −18𝑡+41 Sub in the value of t |𝐴𝐵|= 3(3) 2 −18(3)+41 |𝐴𝐵|= 14 |𝐴𝐵|=3.74 (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 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. 30° X b a 20° X b a 160° 20° X b b 5G

This is the formula for the scalar ‘dot’ product of 2 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: The scalar product can be thought of as ‘the effect of one of the two vectors on the other’ Vector multiplication a b a.b = |a||b| By multiplication 𝒂.𝒃= 𝒂 |𝒃|𝑐𝑜𝑠𝜃 a b θ |a|cosθ 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 This is the formula for the scalar ‘dot’ product of 2 vectors a.b = |a|cosθ|b| By GCSE trigonometry a.b = |a||b|cosθ 5G

Vectors 𝒂.𝒃= 𝒂 |𝒃|𝑐𝑜𝑠𝜃 𝑐𝑜𝑠𝜃= 𝒂.𝒃 𝒂 |𝒃| 5G 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 𝒂.𝒃= 𝒂 |𝒃|𝑐𝑜𝑠𝜃 𝑐𝑜𝑠𝜃= 𝒂.𝒃 𝒂 |𝒃| 5G

This is a way to find the dot product from 2 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: 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… If a = x1i + y1j + z1k and b = x2i + y2j + z2k Then: 𝒂.𝒃= 𝑥 1 𝑦 1 𝑧 1 . 𝑥 2 𝑦 2 𝑧 2 = 𝑥 1 𝑥 2 + 𝑦 1 𝑦 2 + 𝑧 1 𝑧 2 𝒂.𝒃= 𝒂 |𝒃|𝑐𝑜𝑠𝜃 This is a way to find the dot product from 2 vectors 𝑐𝑜𝑠𝜃= 𝒂.𝒃 𝒂 |𝒃| 5G

Use the dot product formula 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: Given that a = 8i – 5j – 4k and b = 5i + 4j – k: a) Find a.b 𝒂.𝒃= 𝑥 1 𝑦 1 𝑧 1 . 𝑥 2 𝑦 2 𝑧 2 Use the dot product formula 𝒂.𝒃= 8 −5 −4 . 5 4 −1 𝒂.𝒃= 𝒂 |𝒃|𝑐𝑜𝑠𝜃 𝒂.𝒃= 8×5 + −5×4 +(−4×−1) 𝒂.𝒃=24 𝑐𝑜𝑠𝜃= 𝒂.𝒃 𝒂 |𝒃| If a = x1i + y1j + z1k and b = x2i + y2j + z2k Then: 𝒂.𝒃= 𝑥 1 𝑦 1 𝑧 1 . 𝑥 2 𝑦 2 𝑧 2 = 𝑥 1 𝑥 2 + 𝑦 1 𝑦 2 + 𝑧 1 𝑧 2 5G

Solve, remembering to use inverse Cos 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: 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 Use the angle formula – you will need to calculate the magnitude of each vector as well… 𝑐𝑜𝑠𝜃= 𝒂.𝒃 𝒂 |𝒃| 𝒂.𝒃= 𝒂 |𝒃|𝑐𝑜𝑠𝜃 |a|= 8 2 + (−5) 2 + (−4) 2 |b|= 5 2 + 4 2 + (−1) 2 |a|= 105 |b|= 42 𝑐𝑜𝑠𝜃= 𝒂.𝒃 𝒂 |𝒃| 𝑐𝑜𝑠𝜃= 𝒂.𝒃 𝒂 |𝒃| Sub in the values If a = x1i + y1j + z1k and b = x2i + y2j + z2k 𝑐𝑜𝑠𝜃= 24 105 42 Then: Solve, remembering to use inverse Cos 𝒂.𝒃= 𝑥 1 𝑦 1 𝑧 1 . 𝑥 2 𝑦 2 𝑧 2 = 𝑥 1 𝑥 2 + 𝑦 1 𝑦 2 + 𝑧 1 𝑧 2 𝜃=68.8° 5G

Vectors 5G 𝒂.𝒃= 𝒂 |𝒃|𝑐𝑜𝑠𝜃 𝑐𝑜𝑠𝜃= 𝒂.𝒃 𝒂 |𝒃| 𝒂.𝒃= 𝑥 1 𝑦 1 𝑧 1 . 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: Given that the vectors a = 2i – 6j + k and b = 5i + 2j + λk are perpendicular, calculate the value of λ. 𝒂.𝒃= 𝑥 1 𝑦 1 𝑧 1 . 𝑥 2 𝑦 2 𝑧 2 Calculate the dot product in terms of λ 𝒂.𝒃= 2 −6 1 . 5 2 λ 𝒂.𝒃= 𝒂 |𝒃|𝑐𝑜𝑠𝜃 𝒂.𝒃= 2×5 + −6×2 +(1×λ) 𝒂.𝒃=−2+λ As the vectors are perpendicular, the dot product must be 0 𝑐𝑜𝑠𝜃= 𝒂.𝒃 𝒂 |𝒃| 0=−2+λ Solve λ=2 If a = x1i + y1j + z1k and b = x2i + y2j + z2k Then: 𝒂.𝒃= 𝑥 1 𝑦 1 𝑧 1 . 𝑥 2 𝑦 2 𝑧 2 Only this value of λ will cause these vectors to be perpendicular… = 𝑥 1 𝑥 2 + 𝑦 1 𝑦 2 + 𝑧 1 𝑧 2 5G

Vectors 5G  Let the required vector be xi + yj + zk 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 dot products of both a and b with the required vector will be 0 𝒂. 𝑥 𝑦 𝑧 =0 𝒃. 𝑥 𝑦 𝑧 =0 and −2 5 −4 . 𝑥 𝑦 𝑧 =0 4 −8 5 . 𝑥 𝑦 𝑧 =0 Given that a = -2i + 5j - 4k and b = 4i - 8j + 5k, find a vector which is perpendicular to both a and b −2𝑥+5𝑦−4𝑧=0 4𝑥−8𝑦+5𝑧=0 Let z = 1 Let z = 1 −2𝑥+5𝑦=4 4𝑥−8𝑦=−5 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 x2 −4𝑥+10𝑦=8 Now solve as simultaneous equations 4𝑥−8𝑦=−5 2𝑦=3 𝑦= 3 2 𝑥= 7 4 So a possible answer would be: 7 4 𝒊+ 3 2 𝒋+𝒌 7𝒊+6𝒋+4𝒌 x4 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 7𝒊+6𝒋+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

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 On this example, the second picture shows the diagram being viewed from the top of the red vector 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 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 y You 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 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 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 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) 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+𝑡𝒃 where t is a scalar parameter 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!) 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 −5 4 𝒃= 7 0 −3 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: This is the position vector we will use This is the direction vector we will use 𝒓=𝒂+𝑡𝒃 where t is a scalar parameter 𝒓=𝒂+𝑡𝒃 3 −5 4 + 𝑡 7 0 −3 𝒓= 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 3 −5 4 + 𝑡 7 0 −3 You need to be able to write the equation of a straight line in vector form (effectively the equation of a 3D line!) 𝒓= Some alternative forms 𝒓=3𝒊−5𝒋+4𝒌+𝑡(7𝒊−3𝒌) 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: (By writing in a different form) 𝒓= 3𝒊+7𝑡 𝒊+(−5)𝒋+(4−3𝑡)𝒌 (By multiplying out the brackets and then re-grouping i, j and k terms) 𝒓=𝒂+𝑡𝒃 where t is a scalar parameter 𝒓= 3+7𝑡 −5 4−3𝑡 (By rewriting again in the original column vector form) 5H

where t is a scalar parameter Vectors You need to be able to write the equation of a straight line in vector form (effectively the equation of a 3D line!) 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= 4 5 −1 𝒃= 6 3 2 Working in 2D – the equation of the line can be calculated by using either: The gradient (direction) and a coordinate (like we just did) Two coordinates (since you can calculate the gradient between them) 3D can also be done either way… Calculating b – a will give you the vector AB, ie) the direction vector that passes through A and B 𝒃−𝒂= 6 3 2 − 4 5 −1 𝒃−𝒂= 2 −2 3 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) 𝒓=𝒂+𝑡(𝒃−𝒂) where t is a scalar parameter Then use (b – a) along with either of the 2 coordinates/position vectors you’re given 𝒓= 4 5 −1 +𝑡 2 −2 3 As you aren’t given the direction vector in this type, you have to work it out by calculating the vector AB (b – a) 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!) 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 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) 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= 3 2 −5 +𝑡 1 −6 −2 3+𝑡=𝑎 2−6𝑡=𝑏 −5−2𝑡=0 𝒓=𝒂+𝑡(𝒃−𝒂) where t is a scalar parameter We can use the bottom equation to find the value of t As you aren’t given the direction vector in this type, you have to work it out by calculating the vector AB (b – a) −5−2𝑡=0 𝑡=−2.5 3+𝑡=𝑎 2−6𝑡=𝑏 3+(−2.5)=𝑎 2−6(−2.5)=𝑏 0.5=𝑎 17=𝑏 The coordinate itself is (0.5, 17, 0) 5H

Vectors 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) You need to be able to write the equation of a straight line in vector form (effectively the equation of a 3D line!) 𝑟= 2 5 −3 +𝑡 6 −2 4 𝑟= 8 3 1 +𝑡 3 −1 2 A vector equation of a straight line passing through the points A and B, with position vectors a and b respectively, is given by: Original vector updated with a different ‘b’ part 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=𝒂+𝑡(𝒃−𝒂) where t is a scalar parameter 𝑟= 2 5 −3 +𝑡 3 −1 2 As you aren’t given the direction vector in this type, you have to work it out by calculating the vector AB (b – a) If t = 2 𝑟= 2 5 −3 + 6 −2 4 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 𝑟= 8 3 1 So a coordinate on the line is (8, 3, 1) 𝑟= 8 3 1 +𝑡 3 −1 2 5H

Teachings for Exercise 5I

Vectors 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) 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! 𝒓= 5𝒊+2𝒋−3𝒌 +𝑡(2𝒊−3𝒋+𝒌) 𝒓= 4𝒊−5𝒋+2𝒌 +𝑠(𝒊−𝒋+6𝒌) 𝒓= 2𝒊−𝒋+2𝒌 +λ(𝟒𝒊−2𝒋−2𝒌) 𝒓= 3𝒊−5𝒋+4𝒌 +𝜇(3𝒊−3𝒋+2𝒌) 5I

Vectors 𝑟= 3 8 −2 +𝑡 2 −1 3 3+2𝑡 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! 8−𝑡 −2+3𝑡 Find the x, y and z coordinates in terms of t and s 𝑟= 7 4 3 +𝑠 2 1 4 7+2𝑠 4+𝑠 3+4𝑠 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 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. Solve simultaneously by making either the t or s terms ‘equal’ 3+2𝑡=7+2𝑠 2𝑡−2𝑠=4 8−𝑡=4+𝑠 −𝑡−𝑠=−4 rearrange 𝑠=1 𝑡=3 −2+3𝑡=3+4𝑠 Sub s and t into the 3rd pair – if it ‘works’ then the lines intersect. If not, then they don’t… −2+3(3)=3+4(1) 7=7 So the lines DO intersect 5I

Vectors 𝑟= 3 8 −2 +𝑡 2 −1 3 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! Sub t = 3 into the first equation and calculate the position vector 𝑟= 3 8 −2 +3 2 −1 3 𝑟= 9 5 7 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= 7 4 3 +𝑠 2 1 4 Sub s = 1 into the second equation and calculate the position vector 𝑟= 7 4 3 +1 2 1 4 𝑟= 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

Vectors You 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 𝑐𝑜𝑠𝜃= 𝒂.𝒃 𝒂 |𝒃| 5J

Vectors You 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 Modulus is used so that you get the acute angle rather than the obtuse one 1 y = Cosθ 90 180 270 360 -1 𝑐𝑜𝑠𝜃= 𝒂.𝒃 𝒂 |𝒃| 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 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 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= 2 2 3 𝑐𝑜𝑠𝜃= 𝒂.𝒃 𝒂 |𝒃| 𝒂.𝒃= 3 −8 −1 . 2 2 3 Calculate the dot product, a.b 𝒂.𝒃= 3×2 + −8×2 +(−1×3) 𝒂.𝒃=−13 |𝒂|= 3 2 + (−8) 2 + (−1) 2 |𝒂|= 74 Calculate the magnitude of a and b |𝒃|= 2 2 + 2 2 + 3 2 |𝒃|= 17 5J

Vectors You 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 𝑐𝑜𝑠𝜃= 𝒂.𝒃 𝒂 |𝒃| Sub in the values we have just calculated 𝑐𝑜𝑠𝜃= −13 74 17 Work out the sum 𝑐𝑜𝑠𝜃= 𝒂.𝒃 𝒂 |𝒃| 𝑐𝑜𝑠𝜃= −0.3665… Since the answer is negative, we need to ‘make it positive’ by multiplying by -1 𝑐𝑜𝑠𝜃=0.3665 𝜃=68.5° 𝒂.𝒃=−13 |𝒂|= 74 |𝒃|= 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