3D Coordinates In the real world points in space can be located using a 3D coordinate system. For example, air traffic controllers find the location a plane by its height and grid reference. z (x, y, z) y O x 1

Write down the coordinates for the vertices x y z A C D E F G H B 1 2 6 (0, 1, 2) (6, 1, 2) (6, 0, 2) (0, 0, 2) (0, 1, 0 ) (6, 1, 0) (0,0, 0) (6, 0, 0) 2

3D vectors are defined by 3 components. For example, the velocity of an aircraft taking off can be illustrated by the vector v. z (7, 3, 2) 2 v y 2 O 3 x 7 3

Any vector can be represented in terms of the i , j and k Where i, j and k are unit vectors (one unit long) in the x, y and z directions. z y O k j x i 4

Any vector can be represented in terms of the i , j and k Where i, j and k are unit vectors in the x, y and z directions. z (7, 3, 2) v y 2 v = ( 7i+ 3j + 2k ) O 3 x 7 5

Magnitude of a Vector A vector’s magnitude (length) is represented by |v| A 3D vector’s magnitude is calculated using Pythagoras Theorem in 3D (3 , 2 , 1) z v y 1 O 2 x 3

Find the unit vector in the direction of u

Addition of vectors

Addition of Vectors For vectors u and v

Negative vector For any vector u

Subtraction of vectors

Subtraction of Vectors For vectors u and v

Multiplication by a scalar ( a number) Hence if u = kv then u is parallel to v Conversely if u is parallel to v then u = kv

CombiningVectors

If z = kw then z is parallel to w Show that the two vectors are parallel. If z = kw then z is parallel to w

The position vector of the point A(3 , 2 , 1) is OA written as a Position Vectors A (3,2,1) z a y 1 O 2 x 3 The position vector of the point A(3 , 2 , 1) is OA written as a

Position Vectors If R is ( 2 , -5 , 1) and S is (4 , 1 , -3) 4 = 1 -3 = 1 -3 2 – -5 1 2 = 6 -4 Then RS is s – r

y component of RS is -5 to 1 = 6 z component of RS is 1 to -3 = -4 Position Vectors as a journey If R is ( 2 , -5 , 1) and S is (4 , 1 , -3) x component of RS is 2 to 4 = 2 y component of RS is -5 to 1 = 6 z component of RS is 1 to -3 = -4 2 RS = 6 -4

Express VT in terms of f, g and h Vectors as a journey Express VT in terms of f, g and h – h VT = VR + RS + ST VT = – h – f + g – f + g

Collinear Points BC = 2AB A is (0 , -3 , 5), B is (7 , -6 , 9) and C is (21 , -12 , 17). Show that A , B and C are collinear stating the ratio AB:BC. BC = 2AB AB and BC are parallel (multiples of each other) through the common point B, and so must be collinear A 1 B 2 C AB : BC = 1 : 2

Collinear Points Given that the points S(–4, 5, 1), T(–16, –4, 16) and U(–24, –10, 26) are collinear, calculate the ratio in which T divides SU. S 3 T 2 U ST : TU = 3 : 2

The journey from Q to R is the same as the journey from P to S Using Vectors as journeys PQRS is a parallelogram with P(3 , 4 , 0), Q(7 , 6 , -3) and R(8 , 5 , 2). Find the coordinates of S. P(3 , 4 , 0) Q(7 , 6 , -3) Draw a sketch The journey from Q to R is the same as the journey from P to S S R(8 , 5 , 2) From P to S S(3 + 1 , 4 – 1 , 0 + 5) S(4 , 3 , 5)

Find the co-ordinates of Q. The point Q divides the line joining P(–1, –1, 0) to R(5, 2 –3) in the ratio 2:1. Find the co-ordinates of Q. P(-1 , -1 , 0) R(5 , 2 , -3) PQ = 2/3 PR Q 1 2 6 PQ = 2/3 3 -3 4 = 2 -2 The journey from P to Q Q(-1 + 4, -1 + 2, 0 – 2) Q(3 , 1 , – 2)

P divides the line joining S(1,0,2) and T(5,4,10) in the ratio 1:3 P divides the line joining S(1,0,2) and T(5,4,10) in the ratio 1:3. Find the coordinates of P. SP = 1/4 ST S(1 , 0 , 2) T(5 , 4 , 10) 4 SP = 1/4 4 8 1 = 1 2 The journey from S to P P 3 1 P(1 + 1, 0 + 1, 2 + 2) P(2 , 1 , 4)

a and b must be divergent, ie joined tail to tail The scalar product The scalar product is defined as being: a . b = |a| |b| cos θ 0 ≤ θ ≤ 180 a θ a and b must be divergent, ie joined tail to tail b

Find the scalar product for a and b when |a|= 4 , |b|= 5 when (a) θ = 45o (b) θ = 90o a . b = |a| |b| cos θ  a . b = 4 × 5 cos 45o  a . b = 20 × 1/√2 × √2/√2  a . b = 10√2 a . b = |a| |b| cos θ  a . b = 4 × 5 cos 90o  a . b = 20 × 0  a . b = 0 When θ = 90o

This equilateral triangle has sides of 3 units. p . q The Scalar Product This equilateral triangle has sides of 3 units. p . q p . q = |p| |q| cos θ  p . q = 3 × 3 cos 60o  p . q = 9 × 1/2  p . q = 41/2 27

This equilateral triangle has sides of 3 units. p . (q + r) The Scalar Product This equilateral triangle has sides of 3 units. p . (q + r) r r p . (q + r)= p . q + p . r 60o  p . q = 3 × 3 cos 60o p and r are not divergent so move r  p . q = 3 × 3 × ½  p . q = 41/2  p . r = 3 × 3 cos 60o  p . r = 9  ½ p . (q + r) = p . q + p . r = 9  p . r = 4½ 28

If a and b are perpendicular then a . b = 0

a . b = a1b1 + a2b2 + a3b3 Component Form Scalar Product a1 b1 If a = a2 and b = b2 a3 b3 a . b = a1b1 + a2b2 + a3b3

Angle between Vectors To find the angle between two vectors we simply use the scalar product formulae rearranged a . b = |a| |b| cos θ a . b = a1b1 + a2b2 + a3b3 a . b |a| |b| a1b1 + a2b2 + a3b3 |a| |b| cos θ = cos θ =

Find the angle between the two vectors below. p = 3i + 2j + 5k and q = 4i + j + 3k 3 p = 2 5 4 q = 1 3 |p| = √(32 + 22 + 52) |q| = √(42 + 12 + 32) |q| = √26 |p| = √38 a1b1 + a2b2 + a3b3 |a| |b| a . b = a1b1 + a2b2 + a3b3 cos θ = = 3×4 + 2×1 + 5×3 29 √38 × √26 = = 0∙923 = 29 θ = cos-1 0∙923 = 22∙7o

Perpendicular Vectors a . b = | a | | b | cosƟ If a and b are perpendicular then a . b = 0 cos 90o = 0 If a and b are perpendicular then a1b1 + a2b2 + a3b3 = 0

If a and b are perpendicular then a . b = 0 Show that a and b are perpendicular 3 1 if a = 2 and b = 2 -1 7 a . b = a1b1 + a2b2 + a3b3  a . b = 3×1 + 2×2 + (-1)×7  a . b = 3 + 4 – 7  a . b = 0  a and b are perpendicular 34

Two properties that you need to be aware of Properties of a Scalar Product Two properties that you need to be aware of a . b = b . a a .( b + c)= a . b + a . c

If | p | = 5 and | q | = 4, find p . (p + q) 60o q p . (p + q) = p . p + p . q = | p | × | p |cos 0o + | p | × | q |cos60o = 5 × 5 × 1 + 5 × 4 × ½ = 25 + 10 = 35 36

Find the acute angle between the diagonals of PQRS T divides PR in the ratio 5:4 P(-2,-1,-4) Show that Q, T and S are collinear, and find the ratio in which T divides QS S(7,2,17) Q(1,5,-7) Find the acute angle between the diagonals of PQRS R(7,8,5) TS = 2QT, vectors are parallel through the common point T, so Q , T , S are collinear T(-2+5,-1+5,-4+5) T(3 , 4 , 1) QT : TS = 1 : 2 37

Find the acute angle between the diagonals of PQRS 38

Hence vectors are perpendicular Vectors u and v are defined by u = 3i + 2j and v = 2i – 3j + 4k Determine whether or not u and v are perpendicular. 3 2 u . v = 2 . -3 0 4  u . v = 3×2 + 2×(-3) + 0×4  u . v = 6 – 6 + 0  u . v = 0 Hence vectors are perpendicular

u . v = 0 if vectors perpendicular For what value of t are the vectors u and v perpendicular ? u . v = 0 if vectors perpendicular t 2 u . v = -2 . 10 3 t  u . v = t ×2 + (-2)×10 + 3×t  u . v = 5t – 20 u . v = 0  5t – 20 = 0 t = 4

VABCD is a pyramid with rectangular base ABCD. The vectors AB, AD and AV are given by AB = 8i + 2j + 2k, AD = -2i + 10j – 2k and AV = i + 7j + 7k Express CV in component form. Look for an alternative route from C to V along vectors you know. Note BC = AD and AB = DC CV = CB + BA + AV 8 AB = 2 2 -2 AD = 10 1 AV = 7 7 CV = -AD – AB + AV 2 -10 = 8 - + 1 7 CV = –5i – 5j + 7k -5 7 =

The diagram shows two vectors a and b, with | a | = 3 and | b | = 22. These vectors are inclined at an angle of 45° to each other. a) Evaluate i) a.a ii) b.b iii) a.b b) Another vector p is defined by p = 2a + 3b Evaluate p.p and hence write down | p |. i) ii) iii) b) p2

Vectors p, q and r are defined by p = i + j – k, q = i + 4k, r = 4i – 3j a) Express p – q + 2r in component form b) Calculate p.r c) Find |r| a) b) c)

hence there are 2 points on the x axis that are 7 units from P The diagram shows a point P with co-ordinates (4, 2, 6) and two points S and T which lie on the x-axis. If P is 7 units from S and 7 units from T, find the co-ordinates of S and T. hence there are 2 points on the x axis that are 7 units from P i.e. S and T and

The position vectors of the points P and Q are p = –i + 3j + 4k and q = 7i – j + 5k respectively. a) Express PQ in component form. b) Find the length of PQ.

120o so, a is perpendicular to b + c Q 60° a b c PQR is an equilateral triangle of side 2 units. Evaluate a.(b + c) and hence identify two vectors which are perpendicular. 120o NB for a.c vectors must diverge ( so angle is 120° ) so, a is perpendicular to b + c Hence

Calculate the length of the vector 2i – 3j + 3k

Find the value of k for which the vectors and are perpendicular Put Scalar product = 0

A is the point (2, –1, 4), B is (7, 1, 3) and C is (–6, 4, 2). If ABCD is a parallelogram, find the co-ordinates of D. The journey B to A is the same as the journey from C to D That journey from C gives: D(–6 + [-5], 4 + [-2], 2 + 1) D(–11, 2 , 3)

b + c is perpendicular to a The vectors a, b and c are defined as follows: a = 2i – k, b = i + 2j + k, c = –j + k a) Evaluate a.b + a.c b) From your answer to (a), make a deduction about the vector b + c a) b) b + c is perpendicular to a

Triangular faces are all equilateral In the square based pyramid, all the eight edges are of length 3 units. Evaluate p.(q + r) Triangular faces are all equilateral

(-1,-3,2) (2,-1,1)  D is (-1 + 9, -3 + 6, 2 – 3)  D is (8, 3, –1) A and B are the points (-1, -3, 2) and (2, -1, 1) respectively. B and C are the points of trisection of AD. That is, AB = BC = CD. Find the coordinates of D (-1,-3,2) (2,-1,1)  D is (-1 + 9, -3 + 6, 2 – 3)  D is (8, 3, –1)

Find the co-ordinates of Q. The point Q divides the line joining P(–1, –1, 0) to R(5, 2 –3) in the ratio 2:1. Find the co-ordinates of Q. P Q R 2 1 (5,2,-3) (-1,-1,0)  Q is (-1 + 4, -1 + 2, 0 – 2)  Q is (3, 1, –2)

VABCD is a pyramid with rectangular base ABCD. VA = – 7i – 13j – 11k, AB = 6i + 6j – 6k, AD = 8i – 4j – 4k ; K divides BC in the ratio 1:3. Find VK in component form. + ¼BC = VA + AB VK + ¼AD = VA + AB = – 7i – 13j – 11k 6i + 6j – 6k 2i – j – k = i – 8j – 18k

AB and BC are scalar multiples, so are parallel. A is the point (2, –5, 6), B is (6, –3, 4) and C is (12, 0, 1). Show that A, B and C are collinear and determine the ratio in which B divides AC AB and BC are scalar multiples, so are parallel. B is common. A, B, C are collinear A B C 2 3 B divides AB in ratio 2 : 3