Vector Product in ijk notation

Consider two vectors, A and B,

We can define a new vector C,

C = A x B
C = A B sin

|C| = |A| |B| sin

Remember, tho', that C is a vector. The direction of C is perpendicular to the plane define by vectors A and B. With the fingers of your right hand, rotate the first vector A into the second vector B. The direction your thumb points is the direction of the new vector C. (In this example, vectors A and B are in the plane of this page or screen and the new vector C points into the page or screen)

With this basic definition of the vector product, what can we do with vectors A and B written in component or i, j, k notation?

A = Ax i + Ay j + Az k

B = Bx i + By j + Bz k

C = A x B

C = [Ax i + Ay j + Az k ] x [Bx i + By j + Bz k]

C = Ax Bx ( i x i ) + Ax By ( i x j ) + Ax Bz ( i x k ) +

+ Ay Bx ( j x i ) + Ay By ( j x j ) + Ay Bz ( j x k ) +

+ Az Bx ( k x i ) + Az Bx ( k x j ) + Ax Bx ( k x k )

What are these cross products of the unit vectors?

 i x i = 0 i x j = k i x k = - j j x i = - k j x j = 0 j x k = i k x i = j k x j = - i k x k = 0

So our cross product becomes

C = Ax Bx ( 0 ) + Ax By ( k ) + Ax Bz ( - j ) +

+ Ay Bx ( - k ) + Ay By ( 0 ) + Ay Bz ( i ) +

+ Az Bx ( j ) + Az Bx ( - i ) + Ax Bx ( 0 )

C = i (Ay Bz - Az By) + j (Az Bx - Ax Bz ) + k (Ax By - Ay Bx)

That is certainly a good enough form to remember and we could stop here. But this expression can also be remembered as a determinant,

| Expanding determinants |

 Vector Product (1) Torque Return to ToC, Ch11, Rolling Motion

(c) Doug Davis, 2001; all rights reserved