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Sep 12th, 2005, 11:46 AM
#1
Thread Starter
Hyperactive Member
Showing dot product is commutative, need help please
Hello everyone. I'm trying to show ab = ba. //communative property of dot product
This is what I have, is it enough to show this?
ab = (ax, ay, az) (bx, by, bz) = axbx + ayby + azbz;
ab = (axi + ayj) (bxi + byj); //note: i and j are unit vectors
= axbx(i)(i) + ayby(j)(j) + axby(i)(j) + aybx(j)(i)
= axbx+ayby
ba = (bx, by, bz) (ax,ay,az) = bxax + byay + bzaz;
ba = (bxi + byj) (axi + ayj)
= bxax(i)(i) + byay(j)(j) + bxay(i)(j) + byax(j)(i)
= bxax + byay
Is this enough to show the dot product is communative?
Any help would be great if it isn't!
C¤ry Sanchez
Computer Science/Engineering
@ Penn State
IBM.zSeries Intern
Mandriva 2007
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Sep 12th, 2005, 03:32 PM
#2
Re: Showing dot product is commutative, need help please
(Note: You should write dot product with a dot, not as normal multiplication.)
If you only work in three dimensions, something like this should be enough:
Code:
a.b
= (ax, ay, az).(bx, by, bz) { defenition of a,b as 3d vector }
= ax bx + ay by + az bz { defenition of dot product }
= ... { multiplication is commutative }
= ...
= b.a
I don't understand where you are going with the unit vectors:
ab = (ax, ay, az) (bx, by, bz) = axbx + ayby + azbz;
ab = (axi + ayj) (bxi + byj);
Where did z go?
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Sep 15th, 2005, 03:24 AM
#3
Re: Showing dot product is commutative, need help please
Even much less is enough:
 Originally Posted by voidflux
ab = (ax, ay, az) (bx, by, bz) = axbx + ayby + azbz;
ba = (bx, by, bz) (ax,ay,az) = bxax + byay + bzaz;
That's simply it! You're being asked to show the dot product is commutative, and this is obvious because the product of 2 numbers is commutative (you're not being asked to prove this!): axbx = bxax (etc), ax and bx being numbers (not vectors).
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