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Figure a: The unit vectors, I hat, j hat and k hat of the x y z coordinate system are shown. Arrows indicate the sequence from I hat to j hat to k hat and back to I hat. Figure b: The unit vectors, I hat, j hat and k hat of the x y z coordinate system are shown. I hat equals j hat cross k hat. j hat equals k hat cross i hat. k hat equals i hat cross j hat. Figure c: The unit vectors, I hat and j hat are shown along with minus k hat pointing down. Minus k hat equals j hat cross i hat. Figure d: The unit vectors, I hat and k hat are shown along with minus j hat pointing to the left. Minus j hat equals i hat cross k hat.
(a) The diagram of the cyclic order of the unit vectors of the axes. (b) The only cross products where the unit vectors appear in the cyclic order. These products have the positive sign. (c, d) Two examples of cross products where the unit vectors do not appear in the cyclic order. These products have the negative sign.

Suppose we want to find the cross product A × B for vectors A = A x i ^ + A y j ^ + A z k ^ and B = B x i ^ + B y j ^ + B z k ^ . We can use the distributive property ( [link] ), the anticommutative property ( [link] ), and the results in [link] and [link] for unit vectors to perform the following algebra:

A × B = ( A x i ^ + A y j ^ + A z k ^ ) × ( B x i ^ + B y j ^ + B z k ^ ) = A x i ^ × ( B x i ^ + B y j ^ + B z k ^ ) + A y j ^ × ( B x i ^ + B y j ^ + B z k ^ ) + A z k ^ × ( B x i ^ + B y j ^ + B z k ^ ) = A x B x i ^ × i ^ + A x B y i ^ × j ^ + A x B z i ^ × k ^ + A y B x j ^ × i ^ + A y B y j ^ × j ^ + A y B z j ^ × k ^ + A z B x k ^ × i ^ + A z B y k ^ × j ^ + A z B z k ^ × k ^ = A x B x ( 0 ) + A x B y ( + k ^ ) + A x B z ( j ^ ) + A y B x ( k ^ ) + A y B y ( 0 ) + A y B z ( + i ^ ) + A z B x ( + j ^ ) + A z B y ( i ^ ) + A z B z ( 0 ) .

When performing algebraic operations involving the cross product, be very careful about keeping the correct order of multiplication because the cross product is anticommutative. The last two steps that we still have to do to complete our task are, first, grouping the terms that contain a common unit vector and, second, factoring. In this way we obtain the following very useful expression for the computation of the cross product:

C = A × B = ( A y B z A z B y ) i ^ + ( A z B x A x B z ) j ^ + ( A x B y A y B x ) k ^ .

In this expression, the scalar components of the cross-product vector are

{ C x = A y B z A z B y , C y = A z B x A x B z , C z = A x B y A y B x .

When finding the cross product, in practice, we can use either [link] or [link] , depending on which one of them seems to be less complex computationally. They both lead to the same final result. One way to make sure if the final result is correct is to use them both.

A particle in a magnetic field

When moving in a magnetic field, some particles may experience a magnetic force. Without going into details—a detailed study of magnetic phenomena comes in later chapters—let’s acknowledge that the magnetic field B is a vector, the magnetic force F is a vector, and the velocity u of the particle is a vector. The magnetic force vector is proportional to the vector product of the velocity vector with the magnetic field vector, which we express as F = ζ u × B . In this equation, a constant ζ takes care of the consistency in physical units, so we can omit physical units on vectors u and B . In this example, let’s assume the constant ζ is positive.

A particle moving in space with velocity vector u = −5.0 i ^ 2.0 j ^ + 3.5 k ^ enters a region with a magnetic field and experiences a magnetic force. Find the magnetic force F on this particle at the entry point to the region where the magnetic field vector is (a) B = 7.2 i ^ j ^ 2.4 k ^ and (b) B = 4.5 k ^ . In each case, find magnitude F of the magnetic force and angle θ the force vector F makes with the given magnetic field vector B .

Strategy

First, we want to find the vector product u × B , because then we can determine the magnetic force using F = ζ u × B . Magnitude F can be found either by using components, F = F x 2 + F y 2 + F z 2 , or by computing the magnitude | u × B | directly using [link] . In the latter approach, we would have to find the angle between vectors u and B . When we have F , the general method for finding the direction angle θ involves the computation of the scalar product F · B and substitution into [link] . To compute the vector product we can either use [link] or compute the product directly, whichever way is simpler.

Practice Key Terms 6

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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