Vectors Revision Notation B The vector AB A

Vectors • Revision

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

Unit Vectors (1) i i is the unit vector in the x-direction i= j [] 1 0 j is the unit vector in the y-direction j= [] 0 1 All vectors can be expressed as a linear combination of these 2 vectors e. g. displacement = [] 25 43. 3 = 25 [ ] + 43. 3 [ ] 1 0 0 1

Unit Vectors (2) i= [] 1 0 j= [] 0 1 All vectors can be expressed as a linear combination of these 2 vectors e. g. displacement = [] 25 43. 3 = 25 [ ] + 43. 3 [ ] 1 0 = 25 i + 43. 3 j This is the standard way displacement vectors are presented 0 1

Magnitude a = -10 i + 15 j a is notation for magnitude -10 i + 15 j a By Pythagoras, the magnitude = √(152 + 102) = √ 325 = 18. 0 (1 d. p. ) a = 18. 0

Magnitude of a 3 D Vector (General) z a t y |a| = √(r 2 + s 2 + t 2) s o “the magnitude is the square root of the sum of the squares of the 3 components. ” r x [Pythagoras in 3 D]

Magnitude of a 3 D Vector - example z A a 10 3 o √(32 + 42) 4 B 3 x By Pythagoras, OB = √(32 + 42) y By Pythagoras, OA 2 = AB 2 + OB 2 AB 2 = 102 OB 2 = (32 + 42) OA 2 = 102 + 32 + 42 OA = √(102 + 32 + 42) |a| = OA = 11. 2

Linear Combinations 3 a + 6 b 2 a - b

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

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

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

Distance between 2 points - general case A and B have position vectors z b What is the distance between them? AB =b-a AB |AB| = √(x 2 -x 1)2 + (y 2 -y 1)2 + (z 22 B y A a x

Distance between 2 points A and B have position vectors z y b B What is the distance between them? a + AB = b - a AB x A |AB| = √(52 + 82) = √ 114 = 10. 7

Parallel Vectors (1) -20 i + 30 j -10 i + 15 j 2 a a -2 i + 3 j Vectors with a scaler applied are parallel 1/ 5 a i. e. with a different magnitude but same direction

Parallel Vectors (2) With linear combinations, look to see if you can rearrange them 2 c - 3 d Is parallel to …. 4 c - 6 d c - 1. 5 d 1000 c - 1500 d -6 c + 9 d 2(2 c - 3 d) 0. 5(2 c - 3 d) 500(2 c - 3 d) -3(2 c - 3 d)

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

Scaler Product a b = a 1 b 1+ a 2 b 2+ a 3 b 3 The angle θ between vectors is given as: cos θ = a b |a||b|

Lets see what you are given

Parallel Vectors 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| If a b = 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

Example (2 D) - angle between vectors Given: a = 3 i + 4 j and The scaler product is written as. . . b = i - 3 j a b = (3 x 1) + (4 x -3) = 3 - 12 = -9 The i components cos θ= a b |a||b| |a| = √(32 + 42) = √ 25 = 5 |b| = √(12 + (-3)2) = √ 10 cos θ = -9 = -0. 569 5√ 10 The j components θ = cos-1(-0. 569) = 124. 7 o

Angle between 3 D Vectors The scaler product is written as. . . a b = (2 x 1) + (3 x -2) + (7 x 5) = 2 - 6 + 35 = 31 cos θ = a b |a||b| cos θ = 31 |a| = √(22 + 32 + 72) = √ 62 |b| = √(12 + (-2)2 + 52) = √ 30 √ 62√ 30 = 0. 719 θ = cos-1(0. 719) = 44. 0 o

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

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

Convert this Cartesian equation into a Vector equation Easiest Method y = 4 x + 3 Write: t = 4 x t=y-3 y - 3 = 4 x = t x= y= 1/ 3 + 4 t t [] [] [ ] x y = r =[ 0 3 +t ] [] 0 3 +t 1 4 1/ 1 4 Can replace with a parallel vector

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

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

Points on line A line has the vector equation r = (i + 2 j - 4 k) + t (3 i + 2 j + 10 k) Show that the point with position vector is on the line 13 i + 10 j + 36 k t that fit all points 1. Rearrange equation A single value of t exists for all points. r = (1+3 t)i + (2+2 t)j + (-4+10 t) Therefore the point is on the line 2. Has to equal 13 i + 10 j + 36 k Need to find a value of 3. Equate coefficients i coefficients 1+3 t = 13 t=4 j coefficients k coefficients 2+2 t = 10 -4+10 t = 36 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 Equating the x values: 1+5 5 for some value of = 11 = 10 =2 Substitute = 2 -3+5 = a -4+8 = b -4+16 = b -3+10 = a b = 12 a=7

Intersect of 2 D lines in vector form - 1 For example [] [] [] x y = 1 +s 3 2 2 and [] [] [] x y 6 +t -2 = -1 4 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 + 2 s = 6 - t y part: 3 + 2 s = -2 + 4 t Subtract x from y : 2 = -8 + 5 t 5 t = 10 t=2 Substitute: 1 + 2 s = 6 - 2 2 s = 3 s = 1. 5 [] [] [] [ ] =[ ]+ [ ]=[ ] x y x y = 1 +1. 5 2 3 2 1 3 3 3 4 6 position vector of the point of intersection

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. 4 x: 2+ y: 3 z: 5+ = 11 3 =6 =2 x : 2 + 4 x 2 = 4 +2 =3 x+y : 5 + Substitute 3 = 4 +2 =7 -2 3 = 2 +3 Check the values in the 3 rd equation z : 5 + 3 x 2 = 2 +3 x 3 11 = 11 Satisfied! Hence a point exists common to both lines

Example (continued) The lines r and s have the equations. . . and Show they intersect and find the point of intersection 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 Substitute The 2 lines intersect at (10, 1, 11) Hence a point exists common to both lines

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

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

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

Skew Angle Example 2 lines have the equations … find the angle between them. r = (2 i + 3 j + 6 k) + t(4 i - j + 6 k) and r = (4 i + 7 j + 8 k) + s(2 i - 2 j + k) cos θ = a b Direction Vectors are: |a||b| a = 4 i - j + 6 k b = 2 i - 2 j + k a b = 4 x 2 + -1 x-2 + 6 x 1 = 16 |a| = √(42 + -12 + 62) = √ 53 |b| = √(22 + -22 + 12) = √ 9 = 3 cos θ = 16 = 0. 733 3√ 53 θ = cos-1(0. 733) = 42. 9 o

Angles Between Skew Lines - you find the angle! between and The direction vectors: and a b = 4 x 2 + -1 x-2 + 3 x 3 = 19 |a| = √(42 + -12 + 32) = √ 26 |b| = √(22 + -22 + 32) = √ 17 cos θ = 19 = 0. 904 √ 26√ 17 θ = cos-1(0. 904) = 25. 3 o