CROSS PRODUCTS! Although we started by looking at a wrench, there are all sorts of instances in physics When we first talked about torque, it was pointed out that the ease with which a nut when we want to product of the magnitude of one vector and the magnitude of the can be rotated by a wrench is going to be related to how big the applied force F is, perpendicular component of the second vector.! where the force acts (defined by a vector r) and how F and r are oriented to one another. Specifically, the amount of umph (torque) the wrench provides is equal to Because it pops up so often, this process is called a cross product. For two vectors C and the distance out “r” times the component of F perpendicular to r. That is:! D, the magnitude of DxC is:! ! ! = r F ! ! ! ! " DxC = C D sin!,

F where ! is the angle between C and D.! F! = F sin" The direction of the cross product will be perpendicular to the plane determined by C and ! D, and can be determined using the right hand rule.! r F ! That means that for cases in which r and F are in the x-y plane, the “direction” of the cross product will be in the + or – k direction.!

This quantity, called torque, is the rotational counterpart to force. That is, if you apply a net force to a body, it will accelerate. By the same token, if you apply a net torque to a body, it will angularly accelerate.!

1.)! 3.)!

F Cross Product in Polar Notation!

F! = F sin" Physically, the cross product between two vectors is itself a vector whose ! direction is perpendicular to the plane defined by the original vectors, and r whose magnitude is equal to the product of the magnitude of one vector and F! the magnitude of the perpendicular component of the second vector. Determine whether the vector is + or – is done using the right-hand rule.! Using the wrench and a little math, we can write:! ! r F ! = " ! ! ! Assume you have two vectors:! 1 = r F sin# !2

! ! ! A = A "! and! B = B "! B This is, in other words, the magnitude of F times the magnitude of r times the sine of the 1 2 angle between the line of F and the line of r.! ! Note that the angle between the two A

vectors is! !1 " !2

3.)! 4.)! Then the magnitude of the cross product of B and A is:! Cross Product in Unit Vector Notation!

! 1 ! ! Assume you have two vectors A and B , where:! ! !2 B ! ! = " # " ˆ ˆ ˆ 1 2 A = A i + A j + A k ! x y z B! = B sin" ! B = B ˆi + B ˆj + B kˆ x y z ! ! ! ! BxA = magnitude of A magnitude of perp. comp. of B ! ( )( ) A ! = A B! ! ! # % = A $ B sin"& ! ! ! = A # B sin('1 ( '2 )% B $ & ! A

5.)! 7.)!

OR! ! 1 ! ! ! We would like to determine! BxA A A ˆi A ˆj A kˆ ! !2 = x + y + z B ! ˆ ˆ ˆ ! = " # " B = Bx i + By j + Bzk ! 1 2 One way to do this would be to execute the A! = A sin" series of mini cross products shown below. ! ! ! ! ! BxA = (magnitude of B)(magnitude of perp. comp. of A) Note that the angle between two vector components in the x-direction is ! ! A zero (hence the first term’s sine value being zero) while the off-direction = B A! ! ! pairs have non-zero cross product whose directions are determined using # % = B $ A sin"& ! ! the right-hand rule.! = B # A sin('1 ( '2 )% $ & ! ! BxA B ˆi B ˆj B kˆ x A ˆi A ˆj A kˆ = ( x + y + z ) ( x + y + z ) In short, the magnitude of the cross product is the magnitude of the one = ! B ˆi x A ˆi # + ! B ˆi x A ˆj # + . . . "( x ) ( x )$ "( x ) ( y )$ vector times the magnitude of the second vector times the sine of the 0! angle between the line of the two vectors, or.! = ! B A sin0o # + ! B A sin 90o # ±kˆ + . . . " x x $ " x y ( )$( ) ! ! ! ! BxA = B A sin!

6.)! 8.)! If you did this series of cross products out to conclusions, you would find To evaluate this matrix, I usually start by reproducing the first and second columns to the three products equal to zero (the products associated with the x-x term, right of the matrix (you will see why shortly)! the y-y term and the z-z term) and six terms (as vectors) that were non- zero. Interestingly, you would get the exact same combination of terms with ˆ ˆ ˆ ˆ i j k ˆi j unit vectors if you just evaluated the matrix shown below. That is, if:! ! ! BxA = Bx By Bz Bx By ! ˆ ˆ ˆ A A = Ax i + Ay j + Azk Ax Ay Az x Ay ! B = B ˆi + B ˆj + B kˆ x y z The beginning of the process is shown below (just for the “i” part). ! then:! ˆ ˆ ˆ ˆ i j k ˆi j ˆi ˆj kˆ ! ! ! ! BxA = Bx By Bz Bx By BxA = Bx By Bz Ax Ay Az Ax Ay Ax Ay Az ˆ = i " By (Az ) ! (Bz ) Ay $ #( ) ( )% Minor note: The first term in the cross product—the B vector—is placed in the second row, and the second term is placed in the third row.! 9.)! 11.)!

To evaluate this matrix, I usually start by reproducing the first and second What’s going on?! columns to the right of the matrix (you will see why shortly)! ˆ ˆ ˆ ˆ i j k ˆi j ˆ ˆ ˆ ˆ ! ! i j k ˆi j BxA B B B B B ! ! = x y z x y BxA = Bx By Bz Bx By Ax Ay Az Ax Ay A A A A A x y z x y ˆ = i " By (Az ) ! (Bz ) Ay $ #( ) ( )% To evaluate this matrix, I usually start by reproducing the first and second columns to the right of the matrix (you will see why shortly)! Blank out the column and row in which exists the unit vector “i.” Multiply that “i” by the evaluation of the little two by two matrix that exists down and to the right (see sketch). !

To evaluate the two by two matrix, you multiply the upper left term by the lower right The nice thing about this operation is that you do the same process over and term, then subtract from that the lower left term times the upper right.! over again, and you don’t have to worry about the cross product’s direction as the math gives you your answer ready-made in u.v.n.! Once done for “I,” you do the same for the “j” terms. !

10.)! 12.)! ! Given:! ˆ ˆ ˆ Adding in the “j” term looks like:! P = 7i ! 4 j + 2k ! ˆ ˆ ˆ Q = !3i ! 5 j + 2k ˆ ˆ ˆ ˆ ! ! i j k ˆi j Determine:! ! ! QxP BxA = Bx By Bz Bx By

Ax Ay Az Ax Ay ˆ ˆ = i " By (Az ) ! (Bz ) Ay $ + j"(Bz )(Ax ) ! (Bx )(Az )$ #( ) ( )% # %

Finishing off with the “k” term gives us:!

ˆ ˆ ˆ ˆ i j k ˆi j ! ! BxA = Bx By Bz Bx By

Ax Ay Az Ax Ay ˆ ˆ ˆ = i " By (Az ) ! (Bz ) Ay $ + j"(Bz )(Ax ) ! (Bx )(Az )$ + k "(Bx ) Ay ! By (Ax )$ #( ) ( )% # % # ( ) ( ) %

13.)! 15.)!

! DON’T MEMORIZE THIS BOTTOM LINE. Instead, learn how to do the process. ! Given:! F 7 30o ! = " EXAMPLE: Determine DxC, assuming that:! o H = 4"150 ! ! ! C = !1 ˆi + 2 ˆj Determine:! ( ) ( ) FxH ! D = 4 ˆi + 5 ˆj + 6 kˆ ( ) ( ) ( )

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ! ! i j k i j ! ! i j k i j ! ! i j k i j DxC = 4 5 6 4 5 +! DxC = 4 5 6 4 5 +!DxC = 4 5 6 4 5 !1 2 0 !1 2 !1 2 0 !1 2 !1 2 0 !1 2

! ! ˆ ˆ ˆ DxC = i #"(5)(0) ! (6)(2)%$ + j#"(6)(!1) ! (4)(0)%$ + k #"(4)(2) ! (5)(!1)%$ ˆ ˆ ˆ = ! 12i ! 6 j + 13k

14.)! 16.)!