Alg1131W4T3 - Cross products

Friday, 23 March 2018

1:29 PM

Machine generated alternative text: The definition in terms of coordinates is al bl The cross product is defined only for three-dimensional vectors, and gives us a way of associating to two vectors a and b a third vector a x b which is perpendicularto both, and whose length is essentially the area of the parallelogram spanned by a and b. The direction of a x b is given in terms of a right hand rule: if the thumb and index fingers of the right hand point in the respective directions of a and b, then the middle finger points in the direction of b ax b. ax b dib2 - "2b1 a2b3 — a3b2 To make this definition a bit easier to remember, we should notice that the indices are transformed via the cyclic rotation 1 —+ 2 —+ 3 —+ 1. For example, applying this to the first entry a2b3 — a3b2 gives the second entry a3b1 So to practice: — 01 b3 2 5 3 4 ii) 2 x 4 0 This second example is particularly instructive; in this case the two vectors actually lie in the xy-plane, and we can see that indeed the cross product is perpendicular to both input vectors If we have a look at the GeoGebra app below, you can check that the parallelogram formed by the vectors u = 6 5 4 3 2 u 3 and v = 2 in the plane has area 10 4 1 2 3 4 5 6

 

Machine generated alternative text: You are given the cross product Let's observe that and 1 0 x 10 10 o 10 This means that 10 is perpendicular ' O to both 1 and 5 O and , the vectors in the given cross product. i) So, using this idea, a (non-zero) vector perpendicular to both ii) Likewise, a (non-zero) vector v satisfying both v • Note: 2 is 0 and v. . recall that u and v are perpendicular precisely when u •v = 0 . make sure to enter your vectors in Maple notation (i.e. < 1, 2, 3

 

Machine generated alternative text: If u , v and w are vectors in three-dimensional space, which of the following are true? (3u) x v 3(u x v)

 

Machine generated alternative text: If a charged particle of charge q is travelling with a velocity v in a magnetic field B, then the force that that charged particle feels is given by In this case; the force F is also a vector quantity since it has both a magnitude and a direction. So the cross product plays an important role in physics and engineering. Now suppose that a proton with some positive charge q is traveling in the my-plane with a velocity in the direction of the vector v 3 —3 and that the magnetic field B is a uniform field pointing straight up in the z direction, perpendicular to the xy-plane Then the direction of the force that the moving proton feels is in the direction <-3, -3, Note: make sure to use Maple notation for the vectors, e.g. < 1, 2, 3 This is a direction which [select all that apply] is parallel to the direction of the magnetic force is perpendicular to the direction of the magnetic force is also in the my-plane is perpendicular to the velocity of the particle.

 

Machine generated alternative text: Challenge: (worth 2 marks) If we observed this particle for a while, we would see that its trajectory was a circle O in the my-plane.

 

Machine generated alternative text: Consider the points A Find the following cross products: . öBxö*A • OC x OA • OA x OC . OD x OC . 0B x OC , 0] and E = [O, —1, 0] lying on the unit circle in the xy-plane_ Furthermore, let u and z Note that the magnitude 10B x OAI is given by the area of the parallelogram spanned by 0B and OA

 

Machine generated alternative text: We know the formula la x bl = lallbll where 9 is the angle between a and b. From this we can deduce that Note: you can enter V'9 by using the notation sqrt (2) More generally if p and q are on the unit circle in the xy-plane, then 1 where r runs through the interval [-1,1] Note: enter the interval using the syntax [1 , m]

 

 

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