A Cross product

A geometrical method of multiplying two vectors is the cross product. The symbol "" designates this as a three-dimensional product method. This product technique has a wide range of uses in physics, engineering, and computer science. Dot and cross products have some similarities because they both use a three-dimensional method, but the dot product is essentially a projection of a vector in a certain direction. From the coordinates or matrix notation, both can be computed. It is restricted to particular algebraic concepts like cross product distribution and scalar multiplication.
Introduction To gather an understanding of cross products, a comparison of the dot and cross products should be established since they tend to be related and in some cases, can be confused. The cross product, which is also the vector product is a binary computation of two vectors in the three dimensional space i.e., two variables that are linearly independent can be crossed, (×). If a and b are the two vectors, then, a × b will generate a vector c, that is perpendicular to both a and b and the normal plane which contains x and y. one of the vectors can be non - linear, i.e. if the two vectors have the same direction, or either of the vectors has a zero length, the vector product generated will be zero i.e. ā × ƃ = 0. An important aspect in the vector product is the fact that if the magnitude of the two vectors is found, its gives the area of the pallalelogram from which the two vectors was generated this is to say that, the magnitude of the product of a and b is the product of the length of a and b- cross product is basically a directed area.
Computation of the cross product in physics can be defined in vectors as a^b though it is rarely used in mathematical calculations to avoid exterior product confusion (Strang, 1991). When computing the vector product, Ө is the angle between the two vectors i.e. ~a and ~b and if the two vectors are parallel, Ө can either be 0 or 180°. The direction of a vector can be given by the right hand thumb rule- point the forefinger of the right hand in the direction of ~a, the middle in ~b the thumb will automatically be the direction of the vector outcome. From the thumb rule, one can also deduce that the vectors are not commutative since if the right hand fingers ae placed in the wrong vector location an opposite direction to the one that was first deduced will result. Another aspect that should be considered in the vector product is the hardness of the coordinate system. The dot product unlike the cross product, is a scalar multiplication of vectors, it is a measure of parallelism. The dot product calculation formula defined with respect to its scaler quantity as ˆa. ˆb = |ˆa|| ˆb| cos Ө and the immediate consequence of the product is the square length i.e. ˆa. ˆa = |ˆa|2. The square of a vector is 1. ˆi, ˆj, and ˆk are used to show the direction of the vectors i.e. coordinates.
While computing both dot and cross products, it is important to note that
i × i = 0 i× j = k i × k = - j
j × i= -k j × j =0 j × k = i
k × i= j k × j = -i k × k= -0
the following two examples can be used to give an understanding of the cross product.
Example 1: if U and V are parallel vectors, show that u × v = 0 (product of any two parallel vectors is zero)
Solution: according to the anti- commutative property of the vector products, u × v = - w × v where u and v are any vectors. If it is deduced that w = v, then v × v = -v × v. if v × v is added on either side of the equation, then when computed, 2v × v = 0 or v × v = 0. The cross product of any vector therefore with itself which is the vector parallel to it, is zero.
From example 1, example 2 can be deduced. Example 2:
Find the standard unit normal for the pairs i, j and j, k and k, i. find the standard unit that is normal to the pairs j, I and k, j and i, k.
Solution
The right hand thumb rule can be applied here i.e., if one curls the fingers of the right hand from i towards j, the direction that the thumb will point to will be the positive z- axis. The standard unit that is normal to pair i, j therefore is k. Applying the same principle on vector j, i, the unit normal can be found to be -k. the standard unit normal to pair k, i, therefore is j and for pair k, j is - i. standard unit normal for i, is - j. from example 1, v × w = 0. For other cases, one may form the unit vector dir. (v × w) = (|| v × w ||)-1 (v × w). Since v × w is perpendicular to vector v and w, dir (v × w) therefor is the standard unit normal for either v, w or w, v. from the other cross product theorems, the length of v × w shows when the equation v × w = 0 holds. It can also be deduced that.
|| v × w ||2 = ||v||2 ||w||2 - (v. w)2 and if v and w are non- zero vectors, then || v × w || = ||v|| ||w|| sin Ө where Ө which is the angle between v and w lies between 0 and π. If v and w were parallel, v × w = 0, if the two vectors were not parallel, then, v w would point in the direction of the standard unit normal for vector pair v, w i.e. dir (v × w) = (||v × w||)-1(v × w) therefore is the standard unit normal for the vector pair v, w. as proof for this, if an identity for v and w are used i.e. v = (v1, v2, v3) and w = (w1, w2, w3), then
||v × w || = (v2 w3-w2v3)2 + (v1w3 - w1v3)2 +(v1w2- w1v2)2
= (v12+ v22 + V32) (w12+ w22 +w32) - (v1w1 + v2w2 + V3w3)2
= ||v||2 ||w||2 - (v.w)2.
The second equality in the chain is not totally obvious and can be verified by multiplying everything. Incase neither v nor w is a zero vector, the scalers generated ||v|| and ||w|| will be non-zero and the identity will therefore become
||v × w||2 = ||v||2 ||w||2- (v. w)2
= ||v||2 ||w||2 (1- {(v. w)2/||v||2 ||w||2})
||v||2 ||w||2(1 - cos 2 (Ө))

Computing the cross product


ā ā │Sin Ө
.

Ө ƃ
The diagram above shows the geometric boundaries and definition of the cross product as denoted by the right hand rule. It can therefore be deduced from the diagram that ā × ƃ = - (ƃ × ā). While computing the cross product of the two vectors, one should be careful not to interchange the two vectors since the cross product has an anti- commutative property. When computing the cross product from the standard orthogonal criteria, the geometric formula will result into the following;
ˆi × ˆj = ˆk
ˆ j × ˆ k = ˆi
ˆk × ˆ i = ˆj
If this concept is to be emphasized using the cyclic nature of the cross product as shown below, using the direction of the angle, the products will get a positive sign i.e. , ˆk .ˆi = ˆj .i
The geometric formula would be reduced to a standard vector product if an orthogonal approach were to be used, (ˆi, ˆj, ˆ k). If ~a = ax ˆi + ay ˆj + a z ˆ k and ~b = b x ˆ ı + b y ˆj + b z ˆ k, then
i j k
ā ×ƃ = ax ax ax
by by by
which can be expanded further as, ~a × ~b= (ax ˆi + ay ˆj + az ˆk) × (bx ˆi + by ˆj + bz ˆ k) = axbx ˆi × ˆI + axby ˆi × ˆj + ... = (aybz − azby) ˆi + (azbx − axbz) ˆj + (axby − aybx) ˆ
If dot product were to be used in place of the cross product, a right handed approach would have been used instead or a thonomal basis would have been used instead. With the cross product, linearity has to be checked to determine if the product can be distributed over addition: b × (~a + ~c) = ~ b × ~a + ~b × ~c. this can be further illustrated as in the diagram below.

When considering the turn in vectors ~a, ~c, and ~a + ~c, the vector product of each of the vector with ~b gives a result that has the same proportion to the projection that is perpendicular to ~b. The solid lines in the figure show the projections. As seen in the diagram, since they are aligned on the plane that is perpendicular to ~b, they can be joined to form a triangle in the middle as shown in the diagram. Two of the vectors that form the sides of the triangle can constitute the third one i.e. the projections of ~a, ~ c, and ~a +~c where the latter is the sum of the first two vectors, but the cross product of each is the rotation of a single side of the triangle that has been rescaled by the length of ~b and the two arrows are perpendicular to the to the prism faces and the two vectors still constitute the third as shown in the diagram where the vector triangle is in front of the prism.
Algebraic properties of the cross product
Distribution, linearity and the Jacobian identity of the cross products show that the R3 orientation in the vector space, and the addition of the vectors and the resultant vector product produce a lie algebra which for the real three dimensional orthogonal group can be represented as SO (3). To show that cross product is distributive over addition, two claims can be made,
(1) c × (a + b) = c × a + c × b
(2) (c + a) × b = c × b + a × b
Another algebraic property of the cross products is that (ka) × b = a × (kb) = k (a × b) where k is any scalar quantity. It can also be noted that a × b is orthogonal to the product of a and b, i.e. the product of the two vectors is perpendicular to the two individual vectors. The cross product is also anti- commutative.


Divergence and curl
Cross products can't be computed without divergence and curl aspects. Divergence is flux per unit volume. It can be easily calculated by taking an infinitesimal amount of the Divergence and it is mostly done to prove the divergence theorem instead of deriving the components of the divergence which can then be represented in rectangular coordinates (Stahler, Clingman & Kahrizi, 2004). This calculation emphasizes on the geometric relationship that exists between the divergence and the flux and also clarifies how to derive component representation in different coordinates. Curl is the circulation per unit area. Circulation of any plane in a coordinate plane is equal to sum of circulations. It can be shown in the diagram below. The triangle has been decomposed into 3 other triangles i.e. three different coordinate planes.

Circulation can then be done with respect to the original triangle as the arrows show in the diagram and the circulations on the sides that are extra can be cancelled pairwise.
Worked examples
Example 3:
Find the cross product of ~a = {1, 2,3} and ~ b = {2, 1, -2}
From the discussion in the paper, cross product of a and b can be calculated using the matrix method i.e.
~a × ~b = i j k
1 2 3 = i (2. (-2) -3. 1) - j (1. (-2) -2. 3) + k (1. 1- 2. 2)
2 1 -2
= i (- 4 - 3) - j (-2 -6) + k (1- 4)
= - 7i + 8j - 3k = {-7, 8, -3}
Example 4
If an object has a weight of 50 N, then it is moved in a vertical plane along a path marked by ABCA where AB = 6 m and AC is 3m as shown in the diagram below.

Determine the amount of work that is done by the gravitational force.
Solution
Work done from point A to B can be determined as below,
F = (0 - 50) N and ~d = AB = (6, 0) m
Since work is the product of force and distance, then, W = F ⋅d = (0)(6) + (−50) (0) = 0J
The work done from B to C can be computed as: F = (0, −50) N and d = BC = (−6,3) m and therefore the work done which is the product can be computed. W = F ⋅d = (0) (−6) +(−50) (3) = −150J
For the work done between C and A, F = (0, −50) N and d = CA = (0, −3) m then computing the work done W = F ⋅d = (0)(0) +(−50) (−3) =150J
The path along ABCA can then be determined since F = (0, −50) N and d = AA = 0 therefore the work done by the gravitational force W = F ⋅d = F ⋅0 = 0J which can also be obtained by summing the work done in the separate sections i.e. A-B, B-C and C-A.

Applications of cross products
Cross products have a number of applications in the engineering, programing and physics field (Szecsei, 2006). Cross products can also be used in a rational motion or product of two vectors e.g., if the torque Ʈ= ~r × ~F, and the Poynting vector ~P = ~E × ~H
= 1/ μ0 (~E × ~B) ... Angular momentum
~L= ~r × ~p, =ώ ×~r,
F = q (~v × ~B)
It can also be used to find angle Ө which can be found as shown below.
Ө = sin-1 [|~A × ~B| / |~A|| ~ B|]
|~A × ~ B| represents the area of a parallelogram whose sides are A and B. the cross product can also be used to calculate the distance between skew lines it is also commonly used in computer graphics to calculate normal for the triangle or polygon.
Suppose in real life, one is riding a bicycle and there is no wind such that the speed of wind is zero, you will feel the breeze in as much as the speed of wind at the moment is zero. this is due to the fact that when riding the bicycle one is moving through the wind. On a day that is windless, the apparent wind will be directed in front and its speed is then equal to the speed of the bicycle. This is a life application of the cross product since the observer is in motion and there is also a relative velocity of the wind that relates to the person riding the bike and is the apparent wind. In the case of roller coasters in an amusement park or in modern par themes, most are designed in such a manner that they respond to the gravitational pull on the earth. There are no engines mounted on the cars and when the train the top of the first slope which is the highest point of the ride, it rolls down and gains speed due to the earths gravitational strength. For the next climb, the speed will be insufficient and the energy can be lost due to friction and the train stops after a while. If there was no friction, it would be impossible for the roller coaster train to stop and it would keep on running provided that no point in the track is higher than the first peak. The vector forces, acceleration and velocity are important to ensure safety of the system. The person designing the system then is then supposed to accurately determine how to balance these factors in order to build a secure system and this is where the cross product comes in since it's the vector calculations that will be used to determine the variants and how to balance them to ensure safety of the system.




















References
Stahler, W., Clingman, D., & Kahrizi, K. (2004). Beginning math and physics for game programmers. Indianapolis: New Riders
Strang, G. (1991). Calculus. Wellesley, Mass: Wellesley-Cambridge Press.
Szecsei, D. (2006). Trigonometry: Homework helpers. Franklin Lakes, N.J: Career Press.






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