Axb Dot bxc x cxa equals a Dot bxc squared How can I prove this using Mathmatica X stands for cross product and dot is or the dot product?
Prove that (axb)n[(bxc)x(cxa)] = [a]n(bxc)]^2 where a,b,and c
are all vectors.
First, multiply out the cross products. Since the cross product
of two vectors is itself a vector, we'll give the cross products
some names to make this a little easier to understand:
(bxc)=(b2c3-b3c2)i-(b1c3-b3c1)j+(b1c2-b2c1)k = vector d
(cxa)=(c2a3-c3a2)i-(c1a3-c3a1)j+(c1a2-c2a1)k = vector v
(axb)=(a2b3-a3b2)i-(a1b3-a3b1)j+(a1b2-a2b1)k = vector u
=> (axb)n[(bxc)x(cxa)] = un[dxv]
(dxv)=(d2v3-d3v2)i-(d1v3-d3v1)j+(d1v2-d2v1)k = vector w
=> un[dxv] = unw = u1w1 + u2w2 + u3w3
Now replace u and w with their vector coordinates (notice that
the negative sign is factored into the middle terms, so the
variables are switched).
u1w1 + u2w2 + u3w3= (a2b3-a3b2)w1 + (a3b1-a1b3)w2 +
(a1b2-a2b1)w3
= (a2b3-a3b2)(d2v3-d3v2) + (a3b1-a1b3)(d3v1-d1v3)+
(a1b2-a2b1)(d1v2-d2v1)
Now we need to expand the v terms back out:
(d2v3-d3v2) = d2(c1a2-c2a1) - d3(c3a1-c1a3) = d2c1a2- d2 c2a1-
d3c3a1 + d3c1a3
(d3v1-d1v3) = d3(c2a3-c3a2) - d1(c1a2-c2a1) = d3c2a3 - d3c3a2-
d1c1a2+ d1c2a1
(d1v2-d2v1) = d1(c3a1-c1a3) - d2(c2a3-c3a2) = d1c3a1 - d1c1a3 -
d2c2a3 + d2c3a2
So: (a2b3-a3b2)(d2v3-d3v2) + (a3b1-a1b3)(d3v1-d1v3)+
(a1b2-a2b1)(d1v2-d2v1) = (a2b3-a3b2)(d2c1a2- d2 c2a1- d3c3a1 +
d3c1a3) + (a3b1 - a1b3)(d3c2a3 - d3c3a2- d1c1a2+ d1c2a1)+
(a1b2-a2b1)(d1c3a1 - d1c1a3 - d2c2a3 + d2c3a2)
= d2c1a2 a2b3- d2 c2a1 a2b3- d3c3a1 a2b3 + d3c1a3 a2b3- d2c1a2
a3b2+ d2 c2a1 a3b2+ d3c3a1 a3b2- d3c1a3 a3b2 + d3c2a3 a3b1 - d3c3a2
a3b1- d1c1a2 a3b1+ d1c2a1 a3b1- d3c2a3 a1b3+ d3c3a2 a1b3+ d1c1a2
a1b3- d1c2a1 a1b3+ d1c3a1 a1b2 - d1c1a3 a1b2 - d2c2a3 a1b2 + d2c3a2
a1b2- d1c3a1 a2b1+ d1c1a3 a2b1+ d2c2a3 a2b1- d2c3a2 a2b1
Some of the terms cancel out, leaving us with;
= d2c1a2 a2b3 - d2 c2a1 a2b3 + d3c1a3 a2b3 - d2c1a2 a3b2 +
d3c3a1 a3b2 - d3c1a3 a3b2 + d3c2a3 a3b1 - d3c3a2 a3b1 + d1c2a1 a3b1
- d3c2a3 a1b3 + d1c1a2 a1b3 - d1c2a1 a1b3 + d1c3a1 a1b2 - d1c1a3
a1b2 + d2c3a2 a1b2 - d1c3a1 a2b1 + d2c2a3 a2b1 - d2c3a2 a2b1
Now factor out d1 , d2 , and d3
= d1(c2a1 a3b1 + c1a2 a1b3 - c2a1 a1b3 + c3a1 a1b2 - c1a3 a1b2 -
c3a1 a2b1) + d2(c1a2 a2b3 - c2a1 a2b3 - c1a2 a3b2 + c3a2 a1b2 +
c2a3 a2b1 - c3a2 a2b1) + d3(c1a3 a2b3 + c3a1 a3b2 - c1a3 a3b2 +
c2a3 a3b1 - c3a2 a3b1 - c2a3 a1b3)
Now we can factor out a dot product of ( d1 + d2 + d3):
= ( d1 + d2 + d3)n[(c2a1 a3b1 + c1a2 a1b3 - c2a1 a1b3 + c3a1
a1b2 - c1a3 a1b2 - c3a1 a2b1) + (c1a2 a2b3 - c2a1 a2b3 - c1a2 a3b2
+ c3a2 a1b2 + c2a3 a2b1 - c3a2 a2b1) + (c1a3 a2b3 + c3a1 a3b2 -
c1a3 a3b2 + c2a3 a3b1 - c3a2 a3b1 - c2a3 a1b3)]
(Remember, to keep from changing the value of the equation we
still need to keep the terms grouped together so that they multiply
by the correct d components.)
Now factor out all the "a" components within the brackets:
= ( d1 + d2 + d3)n[(a1 a3{c2b1 - c1b2} + a1 a2{c1b3 - c3b1} + a1
a1{c3b2 - c2b3}) + (a1 a2{c3b2 - c2b3} + a2 a2{c1b3 - c3b1} + a2
a3{c2b1 - c1b2}) + (a1 a3{c3b2 - c2b3} + a2 a3{c1b3 - c3b1} + a3
a3{c2b1 - c1b2})]
= dn[( a1 a3+ a1 a2+ a1 a1)n({c2b1- c1b2} +{c1b3 - c3b1} +
{c3b2- c2b3}) + (a1 a2 + a2 a2 + a2 a3)n({c1b3- c3b1} + {c2b1-
c1b2} + {c3b2- c2b3}) + ( a1 a3+ a2 a3 + a3 a3)n({c3b2- c2b3} +
{c1b3- c3b1} + {c2b1- c1b2})]
And we know that {c2b1- c1b2} +{c1b3 - c3b1} + {c3b2- c2b3} =
(bxc), so we factor out (bxc):
= dn[(bxc)n[(a1 a3+ a1 a2+ a1 a1) + (a1 a2 + a2 a2 + a2 a3) + (
a1 a3+ a2 a3 + a3 a3)]
= dn[(bxc)n[a1(a3+ a2 + a1) + a2 (a1 + a2 +a3) + a3(a1+ a2 +
a3)]]
= dn[(bxc)n([a1 + a2 +a3]n[a1 + a2 +a3]) = dn[(bxc)n(a n a)]
(from above, remember that d = (bxc) )
= (bxc)n(bxc)n a n a
= [an (bxc)]^2
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