Vectors Revision

( ) Notation 8 5 B The vector AB A … as a column vector 8 across, 5 up

Unit Vectors (1) i i is the unit vector in the x-direction i = [ ] 1 j j is the unit vector in the y-direction j = [ ] 1 All vectors can be expressed as a linear combination of these 2 vectors [ ] = 25 + 43.3 e.g. displacement = 25 43.3 [ ] 1

[ ] = 25 + 43.3 [ ] = 25 i + 43.3 j Unit Vectors (2) i = [ ] j = [ ] 1 j = [ ] 1 All vectors can be expressed as a linear combination of these 2 vectors [ ] = 25 + 43.3 e.g. displacement = 25 43.3 [ ] 1 = 25 i + 43.3 j This is the standard way displacement vectors are presented

a = -10 i + 15 j a a Magnitude is notation for magnitude = 18.0 By Pythagoras, the magnitude = √(152 + 102) = √325 = 18.0 (1 d.p.) a a = 18.0

|a| = √(r2 + s2 + t2) Magnitude of a 3D Vector (General) z a y o x t s “the magnitude is the square root of the sum of the squares of the 3 components.” x [Pythagoras in 3D]

Magnitude of a 3D Vector - example z By Pythagoras, OB = √(32 + 42) A √(32 + 42) a 10 y By Pythagoras, OA2 = AB2 + OB2 AB2 = 102 OB2 = (32 + 42) OA2 = 102 + 32 + 42 OA = √(102 + 32 + 42) 3 o B 4 3 x |a| = OA = 11.2

Linear Combinations 3a + 6b 2a - b

+ = (6i + 4j) + (i - 5j) = 7i - j Linear Combination Example 64 1 7 -5 A car drives from A to B with displacement = 6i + 4j Then he drives from B to C with displacement = i - 5j 6i + 4j B A C i - 5j 1 -5 64 + = 7 -1 The resultant displacement is from A to C (6i + 4j) + (i - 5j) = 7i - j The magnitude of the displacement = √(72 + (-1)2) = √50 = 7.1m (1 d.p.)

3D Vectors Follow all the same rules of 2D Vectors!

Position Vectors R A position vector is r fixed to the origin O A free vector has magnitude and direction, but is not fixed to the origin s

= b - a Distance between 2 points - general case B A A and B have position vectors x y z B b What is the distance between them? A a AB = b - a AB |AB| = √(x2-x1)2 + (y2-y1)2 + (z2-z1)2

a + = b = b - a Distance between 2 points B A |AB| = √(52 + 52 + 82) z A and B have position vectors x y z b B What is the distance between them? AB a A a + = b AB AB = b - a |AB| = √(52 + 52 + 82) = √114 = 10.7 AB

Parallel Vectors (1) -20 i + 30 j -10 i + 15 j -2 i + 3 j 2 a a Vectors with a scaler applied are parallel i.e. with a different magnitude but same direction

Parallel Vectors (2) 2c - 3d 4c - 6d 2(2c - 3d) c - 1.5d 0.5(2c - 3d) With linear combinations, look to see if you can rearrange them 2c - 3d Is parallel to …. 4c - 6d 2(2c - 3d) c - 1.5d 0.5(2c - 3d) 1000c - 1500d 500(2c - 3d) -6c + 9d -3(2c - 3d)

Parallel 3D Vectors =3 c =1/2 c =-2 c Look for scalers of each other Is parallel to …. =3 c =1/2 c =-2 c

Scaler Product ab = a1 b1+ a2 b2+ a3 b3 cos θ = ab The angle θ between vectors is given as: cos θ = ab |a||b|

Lets see what you are given

Parallel Vectors Occur …when cos θ = 1 … so θ = cos-1(1) = 0 degrees cos θ = ab |a||b| Occur …when cos θ = 1 … so θ = cos-1(1) = 0 degrees i.e. the lines are Parallel

Perpendicular Vectors cos θ = a.b |a||b| ab If = 0, …then cos θ = 0 … so θ = cos-1(0) = 90 degrees i.e. the lines are Perpendicular So, if ab = 0 then the lines are perpendicular

Given: a = 3i + 4j and b = i - 3j Example (2D) - angle between vectors The scaler product is written as ... ab ab = (3 x 1) + (4 x -3) = 3 - 12 = -9 The i components The j components |a| = √(32 + 42) = √25 = 5 cos θ= ab |a||b| |b| = √(12 + (-3)2) = √10 cos θ = -9 = -0.569 5√10 θ = cos-1(-0.569) = 124.7o

Angle between 3D Vectors The scaler product is written as ... ab ab = (2 x 1) + (3 x -2) + (7 x 5) = 2 - 6 + 35 = 31 cos θ = ab |a||b| |a| = √(22 + 32 + 72) = √62 |b| = √(12 + (-2)2 + 52) = √30 cos θ = 31 = 0.719 √62√30 θ = cos-1(0.719) = 44.0o

b = pi + qj a = xi + yj E.g. a + tb Vector Equation of a line (2D) y x A line can be identified by a linear combination of a position vector and a free vector y b = pi + qj A parallel vector to line a = xi + yj o x t is a scaler - it can be any number, since we only need a parallel vector E.g. a + tb = (xi + yj) + t(pi + qj)

[ ] [ ] y = 3x - 1 r is the position vector of r Often written ……. Equations of straight lines y = 3x - 1 ….. is a Cartesian Equation of a straight line x y [ ] = + t 1 3 2 ….. is a Vector Equation of a straight line Often written ……. = + t 1 3 [ ] 2 r r is the position vector of any point R on the line Any point Direction

[ ] [ ] [ ] r y = 4x + 3 y - 3 = 4x = t = + t = + t Convert this Cartesian equation into a Vector equation Easiest Method y = 4x + 3 y - 3 = 4x = t Write: t = 4x t = y - 3 x = 1/4 t y = 3 + t x y [ ] = + t 3 1/4 1 [ ] Can replace with a parallel vector = + t 1 4 [ ] 3 r

Equations of form y-b=m(x-a) Summary A line can be identified by a linear combination of a position vector and a free [direction] vector = + t 1 m [ ] a b r the direction vector Any point Equations of form y-b=m(x-a) Line goes through (a,b) with gradient m

r a + = b = b - a Equation given 2 points B A z b AB a AB x AB AB y z A and B have position vectors b B AB What is the vector equation of the line passing through them a A a + = b AB AB = b - a = + t r The equation of the line AB This gives the direction vector

r r = (1+3t)i + (2+2t)j + (-4+10t) = + t 13i + 10j + 36k Points on line A line has the vector equation (3i + 2j + 10k) (i + 2j - 4k) = + t r 13i + 10j + 36k Show that the point with position vector is on the line Need to find a value of t that fit all points 1. Rearrange equation A single value of t exists for all points. Therefore the point is on the line r = (1+3t)i + (2+2t)j + (-4+10t) 13i + 10j + 36k 2. Has to equal i coefficients j coefficients k coefficients 3. Equate coefficients 1+3t = 13 2+2t = 10 -4+10t = 36 t = 4 t = 4 t = 4

Example = + A line has the equation ... The point P (11, a, b) is on the line. a and b are constants. = + Find a and b P has position vector Therefore for some value of Substitute = 2 -3+5 = a -4+8 = b Equating the x values: 1+5 = 11 -3+10 = a -4+16 = b 5 = 10 a = 7 b = 12 = 2

Intersect of 2D lines in vector form - 1 For example x y [ ] 1 3 [ ] 2 [ ] x y [ ] [ ] [ ] 6 -2 -1 4 = + s and = + t If the lines intersect, there must be values of s and t that give the position vector of the point of intersection. x part: 1 + 2s = 6 - t y part: 3 + 2s = -2 + 4t x y [ ] = +1.5 1 3 2 x y [ ] = + 1 3 Subtract x from y : 2 = -8 + 5t 5t = 10 t = 2 [ ] x y = 4 6 position vector of the point of intersection Substitute: 1 + 2s = 6 - 2 2s = 3 s = 1.5

Example The lines r and s have the equations ... Show they intersect and find the point of intersection and If the lines intersect, there must be values of and  that give the position vector of the point of intersection. x : 2 + 4 = 4 +2 y : 3 - = 7 - 2 z : 5 + 3 = 2 +3 x+y : 5 + 3 = 11 3 = 6 = 2 Check the values in the 3rd equation z : 5 + 3x2 = 2 +3x3 11 = 11 Satisfied! Substitute x : 2 + 4x2 = 4 +2 = 3 Hence a point exists common to both lines

Example (continued) The lines r and s have the equations ... Show they intersect and find the point of intersection and If the lines intersect, there must be values of and that give the position vector of the point of intersection. = 2 = 3 Values satisfy all equations Hence a point exists common to both lines Substitute The 2 lines intersect at (10, 1, 11)

Skew lines z b a x In 3D lines can be that are not parallel and do not intersect are called skew lines x y z Don’t meet b a

Skew Example r = (2i + 3j + 6k) + t(4i - j + 6k) 2 lines have the equations ... r = (2i + 3j + 6k) + t(4i - j + 6k) Show they are skew and r = (4i + 7j + 8k) + s(2i - 2j + k) If the lines intersect, there must be values of s and t that give the position vector of the point of intersection. i : 2 + 4t = 4 +2s Check the values in the 3rd equation j : 3 - t = 7 - 2s k : 6 + 6x2 = 8 + 3 18 = 11 k : 6 + 6t = 8 + s Not Satisfied! i+j : 5 + 3t = 11 3t = 6 t = 2 Direction vectors: (4i - j + 6k) and (2i - 2j + k) are not parallel Therefore lines are skew Substitute i : 2 + 4x2 = 4 +2s s = 3

Angles Between Skew Lines Skew lines do not meet! However you can work out angle between them by ‘transposing’ one to the other - keeping the direction the same. E.g. the angle between and You just need to look at the angle between the direction vectors: and

cos θ = ab Skew Angle Example |a||b| 2 lines have the equations … find the angle between them. r = (2i + 3j + 6k) + t(4i - j + 6k) and r = (4i + 7j + 8k) + s(2i - 2j + k) cos θ = ab |a||b| Direction Vectors are: a = 4i - j + 6k b = 2i - 2j + k ab = 4x2 + -1x-2 + 6x1 = 16 |a| = √(42 + -12 + 62) = √53 |b| = √(22 + -22 + 12) = √9 = 3 cos θ = 16 = 0.733 θ = cos-1(0.733) = 42.9o 3√53

Angles Between Skew Lines - you find the angle! and The direction vectors: and ab = 4x2 + -1x-2 + 3x3 = 19 |a| = √(42 + -12 + 32) = √26 |b| = √(22 + -22 + 32) = √17 cos θ = 19 = 0.904 √26√17 θ = cos-1(0.904) = 25.3o