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[GENERAL MATHS A CHEAT SHEET] Unit 2, 2007

Stephen Machet 2007

Matrices

3 2 Matrix = + + 2 2 2 1 2 1

= For , = (Determinant)f = ,1 = 1

For two matrices, and ; + = +

=

(

)

= + = + =EX: Solve for and .3 2 = 55 3 = 93 2

5 3 = 59Let =

1

=

1

=1 =1This finds the coordinates of the intersection.

Vectors

Vectors have magnitude and direction.

Position vectors start at the origin.

= 23

=Magnitude (Length) of = 2 + 2 = 4 + 9 = 13Located vectors: starts at, ends at .Opposite vectors: Parallel vectors which point in opposite

directions.

Negative of a vector: Same magnitude, point in opposite

direction.

Equivalent vectors: Parallel, same magnitude, same direction.Collinear: Same gradient, start at same point.

EX: Show that the points(1,2), (2,4), and (3,6) arecollinear. =

= 24 1

2

= 12

=

= 36 12= 2

4

= 12

The two lines are parallel.Since and start at the same point (), the points, ,and are collinear.Coplanar vectors: Vectors that lie on the same plane 2 non-

zero, non-parallel vectors.

EX: Let = 21, = 12. Express the vector = 31 interms of a linear combination a and b. = + 31 = 2

1 + 1

2

3 = 2 + And 1 = + 2Solve for and , and then substitute into .

=

Vectors in 3D

= = + +

Trigonometric Functions

EX: Ifsin = 47, find cos where

2< <

By trig: 42 + 2 = 72 = 33

cos

=

33

7

Remembering cos is negative in Q2sin cos = 1

2sin2

2sin cos = sin6

3 + 3 + 2

47

(2,3)

Test: and and and Until there is a relation.

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[GENERAL MATHS A CHEAT SHEET] Unit 2, 2007

Stephen Machet 2007

Kinematics

Acceleration due to gravity = 9.81 Formulae for StraightLine Motion = + = 1

2 +

=

+

1

22

= 1222 = 2 + 2

GradientArea

under

Graph

7/27/2019 5394GMA2Cheat Sheet

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[GENERAL MATHS A CHEAT SHEET] Unit 2, 2007

Stephen Machet 2007

Statics

1 = 9.8Triangle of forces:

1sin = 2sin = 3sinTHIS ONLY WORKS FOR THREE ANGLES!

Cos Rule: 2 = 2 + 2 2 cosSin Rule:

cos =

sin =

sin = 2

Resolution of Forces

THIS WORKS FOR MANY ANGLES/VECTORS!

= sin + cosis along the line of possible motion.

is perpendicular to .Complex Numbers

Polar form: = cis = 2 + 2 = tan1 = cos = sinCartesian form: = + Re(z) is m(z) is f = 2 3

Then = 2 + 3And = 2 3 and so onArgand Diagrams

+ = 2 = 2 = 2 + 2 = = + = cos + sin = cos + sin = cis is the modulus of (ie ||), the polar angle, is the argument of z (ie arg )Principle value of the argument (Arg ) occurs for < = cis = cis()If1 = 1cis1 and 2 = 2cis2Then 12 = 12cis(1 + 2)And

12 =12 cis(1 2)

De Moivres Theorem = cis = (cos + sin)

= 1 = = ()

= + Im(

)

cos sin

180 180 1

2

3

3

1 2

180

Re

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[GENERAL MATHS A CHEAT SHEET] Unit 2, 2007

Stephen Machet 2007

Loci