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Advanced Math/linear algebra parallelogram and cross prodcut problem


I have no idea what to do for this problem. Any help would be appreaciated.

Consider the parallelogram of the following picture.

a) Rewrite the following cross products using (vector v) cross product (vector w). Simplify your answers.

i)   (vector AB) cross product (vector AC)
ii)  (vector AC) cross product (vector AD)
iii) (vector AB) cross product (vector DC)
iv)  (vector AC) cross product (vector BD)

(b) For each of the following pairs of vectors, draw the original parallelogram and a new parallelogram whose sides are equal to the given vectors. Use these parallelograms to explain geometrically the corresponding algebraic result of (a).

i)   vector AB, vector AC
ii)  vector AC, vector AD
iv)  vector AC, vector BD

Let's first get some notation defined:

AB = vector A->B = V is your figure
|AB| = magnitude (length) of AB
ABxDC = cross product of vectors AB and AD
|ABxDC| = |AB||DC|sin(a) where a is the angle between the vectors.
AD = W

The key properties of the cross product used to solve these problems are (for vectors V and W):

VxW = U = vector perpindicular to V and W with |U| = |V||W|sin(a)
|U| = area of parallelogram formed by vectors V and W (as in your figure)
Xx(V+W) = XxV+XxW (distributive in addition)
VxW = -WxV.

For Part a):

I: AC is a diagonal of the parallelogram and lies in the same plane as V and W (as does AB). Therefore, ABxAC is perpindicular to V and W. Note that AC bisects the paralellogram and forms a triangle with V. It can be seen that the parallelogram formed by  AC and V has half the area of the parallelogram formed by V and W. Forming the paralellogram of V with AC creates the same triangle so |ACxV| = area of 2 triangles = area of parallelogram of V and W. So ADxAC = VxW.

II: Similar reasoning gives ACxAD = -VxW, where the minus sign comes from the fact that the angle between AD and AC is in the opposite sense as in V and W (the angle is negative).

III: AB and DC are parallel so that ABxDC = 0 (sin(0) = 0)

IV: For the cross-product of the diagonals, note that, from vector addition rules, BD - AB - AD and that AC = AD + DC. Using the distributive property

ACxBD = (AD+DC)x(AB-AD) = ADxAB - ADxAD + DCxAB - DCxAD.

Right off the bat, we have ADxAD = DCxAB = 0 since the vector in the product are parallel. Now, ADxAB = -ABxAD = -VxW. Also, -DCxAD = ADxDC = |AD||DC|sin(b) = |AD||DC|sin(180-a) = |V||W|sin(a) by trig identities and the fact that the angle between the vector AD and DC is the complement of the angle between V and W. Putting this together gives ACxBD = 2・VxW.

You should check to see that I got all the minus signs correct (and everything else). This seems like a pretty standard and important result so it deserves verify and/or finding the result in the literature (i.e. Google).

Part b):

Given the above definitions and derivations, you should be able to plot and interpret the results fairly easily. For IV, a little bit of creative triangle matching should give the result.

Let me know if this makes sense.


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randy patton


college mathematics, applied math, advanced calculus, complex analysis, linear and abstract algebra, probability theory, signal processing, undergraduate physics, physical oceanography


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