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2.5.3 Cross Product

• First, consider an example,

F = (– 6.43i + 7.66j + 0k )N

d = (2i + 0j + 0k )m

M = d ×F =		i	j	k
		2m	0m	0m
	–6.43N 7.66N 0N

NOTE: note that the cross product here is for the right hand rule coordinates. If the left handed coordinate system is used F and d should be reversed.

M = (0m0N – 0m(7.66N ))i–(2m0N – 0m(–6.43N ))j + (2m(7.66N )– 0m(–6.43N ))k = 15.3k(mN )

NOTE: there are two things to note about the solution. First, it is a vector. Here there is only a z component because this vector points out of the page, and a rotation about this vector would rotate on the plane of the page. Second, this result is positive, because the positive sense is defined by the vector system. In this right handed system find the positive rotation by pointing your right hand thumb towards the positive axis (the ‘k’ means that the vector is about the z-axis here), and curl your fingers, that is the positive direction.

• The basic properties of the cross product are,

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The cross (or vector) product of two vectors will yield a new vector perpendicular to both vectors, with a magnitude that is a product of the two magnitudes.

V1 ×V2

V1

V1 ×V2 = (x	1i + y1j + z1	k	)×(x2i + y	2j + z2	k	)					V2

i j k

V1 ×V2 = x1 y1 z1

x2 y2 z2

V1 ×V2 = (y1z2 – z1y2 )i + (z1x2 – x1z2 )j + (x1y2 – y1x2 )k

We can also find a unit vector normal ‘n’ to the vectors ‘V1’ and ‘V2’ using a cross product, divided by the magnitude.

λ n =	V1	×V2
	--------------------
	V1	×V2

• When using a left/right handed coordinate system,

The positive orientation of angles and moments about an axis can be determined by pointing the thumb of the right hand along the axis of rotation. The fingers curl in the positive direction.

y

x

z

x

z

y

+

+

+

z

y

x

• The properties of the cross products are,

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The cross product is distributive, but not associative. This allows us to collect terms in a cross product operation, but we cannot change the order of the cross product.

r1 ×F + r2 ×F = (r1 + r2 )×F	DISTRIBUTIVE
r ×F ≠ F ×r	NOT ASSOCIATIVE
but
r ×F = –(F ×r )

2.5.4 Triple Product

When we want to do a cross product, followed by a dot product (called the mixed triple product), we can do both steps in one operation by finding the determinant of the following. An example of a problem that would use this shortcut is when a moment is found about one point on a pipe, and then the moment component twisting the pipe is found using the dot product.

	(d ×F )•u =	ux uy uz
		dx dy dz
		Fx Fy Fz

2.5.5 Matrices

•Matrices allow simple equations that drive a large number of repetitive calculations - as a result they are found in many computer applications.

•A matrix has the form seen below,

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n columns

										If n=m then the matrix is said to be square.
	a11		a21	…	an1					Many applications require square matrices.
										We may also represent a matrix as a 1-by-3
				…
	a12		a22		an2
								m	rows	for a vector.
	…		… … …
a1m a2m …					anm

• Matrix operations are available for many of the basic algebraic expressions, examples are given below. There are also many restrictions - many of these are indicated.

	B =	3		4	5		C =	12		13	14		D =	21			E =
A = 2		6		7	8			15		16	17			22				24 25 26
		9		10 11				18		19	20			23

Addition/Subtraction

3 + 2

4 + 2 5 + 2

B + C =

3 + 12

4 + 13

5 + 14

A + B =

6 + 2

7 + 2 8 + 2

6 + 15

7 + 16

8 + 17

9 + 2

10 + 2 11 + 2

9 + 18

10 + 19 11 + 20

B + D =

not valid

B – A =

3 – 2

4 – 2

5 – 2

B + C =

3 – 12

4 – 13

5 – 14

6 – 2

7 – 2

8 – 2

6 – 15

7 – 16

8 – 17

9 – 2

10 – 2 11 – 2

9 – 18

10 – 19 11 – 20

B – D =

not valid

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Multiplication/Division
											3		4	5
			3(2 ) 4(2 ) 5(2 )
											-- -- --
A B =			6(2 ) 7(2 ) 8(2 )								2		2	2
									B	=	6		7	8
			9(2 )10(2 )11(2 )						--		--		--	--
									A		2 2 2
											9		10 11
											-- ----- -----
											2		2	2
B D =				(3 21 + 4 22 + 5 23 )
				(6 21 + 7 22 + 8 23 )					D E = 21 24 + 22 25 + 23 26
	(9 21 + 10 22 + 11 23 )

(3 12 + 4 15 + 5 18 ) (3 13 + 4 16 + 5 19 ) (3 14 + 4 17 + 5 20 ) B C = (6 12 + 7 15 + 8 18 ) (6 13 + 7 16 + 8 19 ) (6 14 + 7 17 + 8 20 )

(9 12 + 10 15 + 11 18 )(9 13 + 10 16 + 11 19 )(9 14 + 10 17 + 11 20 )

B B D

---,--- ,--- etc, = not allowed (see inverse)

C D B

Note: To multiply matrices, the first matrix must have the same number of columns as the second matrix has rows.

Determinant
													7 8			6 8			6 7
	B				= 3										– 4		+ 5			= 3 (–3 )– 4 (–6 )+ 5 (–3 ) = 0
													10 11			9 11			9 10
														= (7 11 )– (8 10 ) = –3
					7 8
					10					11
										8				= (6 11 )– (8 9 ) = –6
						6
						9				11
										7				= (6 10 )– (7 9 ) = –3
						6
						9				10
					D			,		E			= not valid (matrices not square)

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Transpose

			3			6	9
		T							DT =							24
	B		= 4			7	10							T
										21 22 23			E		=
																25
			5			8	11
																26

Adjoint

																8			–		6			8				6	7				T
													7																				The matrix of determinant to
													10			11					10 11							9 10
																																	the left is made up by get-
	B						=				–			4		5					3 5					–		3	4				ting rid of the row and col-
																																	umn of the element, and
														10 11							9 11							9	10
																																	then finding the determi-
															4	5				–			3 5					3	4				nant of what is left. Note
																																	the sign changes on alter-
															7	8							6 8					6	7				nating elements.
		D								=	invalid (must be square)

Inverse

x

D = B y

z

B	–1	=	B
			--------
			B

To solve this equation for x,y,z we need to move B to the left hand side. To do this we use the inverse.

x B–1D = B–1 B y

					z
B–1D = I	x			=		x
	y					y
	z					z

In this case B is singular, so the inverse is undetermined, and the

matrix is indeterminate.

D–1 = invalid (must be square)