110 cross product

√42+ 52+ (–12)2

cross product (vector product) Many problems in

three-dimensional geometry require physicists and

mathematicians to find a

VECTOR

that is perpendicular

to each of two given vectors aand b. The cross prod-

uct, denoted a×b, is designed to be such a vector.

One begins by positioning the two vectors aand b

at the same point in space. These vectors define a plane

in space, and one sees that there are two possible direc-

tions for a third vector to point so as to be perpendicu-

lar to this plane. Mathematicians have settled on the

convention of following the “right-hand rule” to deter-

mine which direction to choose:

Take your right hand and point your fingers in

the direction indicated by the first vector a.

Now orient your hand, with your fingers still

pointing in this direction, in such a way that

your palm faces to the side of the plane con-

taining the vector b. (If you curl your fingers,

they will consequently turn through the small-

est angle that leads from ato b.) The direction

in which your thumb now points is the direc-

tion the vector a×bwill take.

Thus a×bwill be a vector that points in one direction

while b×awill point in the opposite direction. (In fact:

b×a= –a×b.)

Mathematicians have settled on a second conven-

tion to define the magnitude of a×b:

The magnitude of a×bis the area of the

PAR

-

ALLELOGRAM

defined by the vectors aand b.

If θis the smallest angle between aand b, then the par-

allelogram defined by the two vectors has side-lengths

|a| and |b|. Taking |a| as the base, the height of the par-

allelogram is then given by |b|·sinθ, and consequently

the

AREA

of the parallelogram is |a|·|b|·sinθ. Thus:

a×bis defined to be the vector of magnitude

|a|·|b|·sinθwith direction given by the right-

hand rule.

For example, if i= <1,0,0> is the unit vector pointing

in the direction of the x-axis, and j= <0,1,0> the unit

vector in the direction of the y-axis, then i×jis a vec-

tor pointing in the direction of the z-axis, with length

equal to the area of the unit square defined by iand j,

namely 1. Thus:

i×j= <0,0,1> = k

If two vectors aand bare parallel, then the angle

between them is zero and a×b= 0.

There is an alternative method for computing cross

products. If ais given by a= <a1,a2,a3> = a1i+ a2j+ a3k

and bis given by b= <b1,b2,b3> = b1i+ b2j+ b3k, then

one can check that the

DOT PRODUCT

of the vector:

(a2b3– a3b2)i+ (a3b1– a1b3)j+ (a1b2– a2b1)k

with each of aand bis zero. Thus this new vector is

perpendicular to both aand b. Mathematicians have

shown that it also has direction given by the right-hand

rule and magnitude equal to the parallelogram defined

by aand b. Thus this new vector is indeed the cross

product of aand b:

where, for the final equality, we have written the for-

mula in terms of the

DETERMINANT

of a 3 ×3 matrix.

(To prove that this new vector does indeed match the

quantity a×b, rotate the system of vectors a, b and

a×bso that apoints in the direction of the x-axis and

blies in the xy-plane. Then, for the rotated system we

have: a1= |a|, a2= 0, a3= 0, b1= |b|cosθ, b2= |b|sinθ,

b3= 0. One can now readily check that the formula

above yields a vector of the required length |a|·|b|·sinθ

pointing in the correct direction. One then argues that

the formula continues to hold when the system of three

vectors is rotated back to its original position.) Thus,

for example, if a= <1,4,2> and b= <3,0,1>, then a×b

= <4·1 – 2·0, 2.3 – 1·1, 1·0 – 4·3> = <4,5, – 12>.

According to the

DISTANCE FORMULA

, this vector has

length = √

–

185, which must be the

area of the parallelogram formed by aand b.

In two-dimensions, the determinant

gives a vector the same length as

a= <a1,a2> and perpendicular to it. In four-dimensional

space one can always find a fourth vector perpendicu-

lar to each of any given three vectors.

See also

ORTHOGONAL

;

TRIPLE VECTOR PRODUCT

;

VECTOR EQUATION OF A PLANE

.