Vector Products - Transformations and Projections in Computer Graphics - page 218

Exercise A.1: Show it!
Perhaps the best proof is to construct the cross product from first principles. Given
the two vectors P and Q , we are looking for a vector R perpendicular to both P and Q .
This requirement does not fully define R ,sinceboth R and − R satisfy it and since it

says nothing about the magnitude of R . We therefore extend our definition of the cross
product by requiring that the triplet ( P , Q , R ) be a right-handed triad of vectors and
also that the magnitude of R be the product

| P || Q | sin θ ,where θ is the angle between

P and Q .
The derivation exploits the orthogonality of the three coordinate axes i =( 1 , 0 , 0 ),
j =( 0 , 1 , 0 ), and k =( 0 , 0 , 1 ) and also uses our definition. The definition implies that
i

i = 0 because the angle between i and itself is zero, and the same holds for j and k .It
also implies that the cross product of any two of the three basis vectors is a unit vector
because the basis vectors are unit vectors and because sin 90 ◦ =1. Oncewerealize
that the triplet ( i , j , k ) is a right-handed triad, we can deduce the following: i

×

× j = k ,

j × k = i , k j .

Armed with this information, we can easily derive the cross product R :

× i = j , j × i = − k , k × j = − i ,and i × k = −

R = P × Q =( P 1 i + P 2 j + P 3 k ) × ( Q 1 i + Q 2 j + Q 3 k )

=( P 1 i + P 2 j + P 3 k ) × Q 1 i +( P 1 i + P 2 j + P 3 k ) × Q 2 j +( P 1 i + P 2 j + P 3 k ) × Q 3 k

=( P 2 Q 3 − P 3 Q 2 ) i +( − P 1 Q 3 + P 3 Q 1 ) j +( P 1 Q 2 − P 2 Q 1 ) k

=( P 2 Q 3 − P 3 Q 2 , − P 1 Q 3 + P 3 Q 1 ,P 1 Q 2 − P 2 Q 1 ) .

Themagnitudeof R can be calculated explicitly

2 =( P 2 Q 3 −

P 3 Q 2 ) 2 +( P 1 Q 3 + P 3 Q 1 ) 2 +( P 1 Q 2 − | R | − P 2 Q 1 )

=( P 1 + P 2 + P 3 )( Q 1 + Q 2 + Q 3 ) ( P 1 Q 1 + P 2 Q 2 + P 3 Q 3 ) 2

−

2 2

( P · Q ) 2 =

2 2

cos θ ) 2

= | P | | Q | − | P | | Q | − ( | P || Q |

2 2 (1 − cos 2 θ )= | P | 2 2 sin 2 θ.

= | P | | Q | | Q |

To illustrate the magnitude, we can draw the parallelogram defined by P and Q (with
an angle θ between them) and show that vector Q sin θ is perpendicular to P .
The following expressions show how P

× Q can be expressed by means of a deter-

minant,

= i

− j

+ k

i j k

P 2 P 3 P 1 P 3 P 1 P 2

P × Q = P 1 P 2 P 3

Q 2 Q 3 Q 1 Q 3 Q 1 Q 2

Q 1 Q 2 Q 3

=( P 2 Q 3 − P 3 Q 2 , − P 1 Q 3 + P 3 Q 1 ,P 1 Q 2 − P 2 Q 1 ) ,

or,alternatively,bymeansofamatrix

⎛ ⎞

0 P 3 − P 2

⎝ ⎠ .

P × Q =( Q 1 ,Q 2 ,Q 3 ) − P 3 0 P 1 (A.5)

P 2 − P 1 0