Cross Products

  • Thread starter Neen87
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  • #1
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Homework Statement



Consider the vector equation a × x = b in R3, where a doesn't = 0. Show that:
(a) a · b = 0
(b) x = (b × a / ||a||^2) + ka is a solution to the equation, for any scalar k

Homework Equations



I'm not really sure, but I've been messing with these:

1. u x (v x w) = (u · w)v - (u · v)w
2. v x w = v2w3 - v3w2, v3w1 - v1w3, v1w2 - v2w1


The Attempt at a Solution



(a) for a · b = 0, b must = 0 since a can't = 0. for b = 0, x must = 0 since a can't = 0.

(b) i'm really lost!


x = ((a x X) x a / (a · a)) + ka

I have no idea if I'm headed in the right direction or where to go with this. Even a few hints would be greatly appreciated!

Thanks!
Tina
 
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Answers and Replies

  • #2
LCKurtz
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Homework Statement



Consider the vector equation a × x = b in R3, where a doesn't = 0. Show that:
(a) a · b = 0
(b) x = (b × a / ||a||^2) + ka is a solution to the equation, for any scalar k

Homework Equations



I'm not really sure, but I've been messing with these:

1. u x (v x w) = (u · w)v - (u · v)w
2. v x w = v2w3 - v3w2, v3w1 - v1w3, v1w2 - v2w1


The Attempt at a Solution



(a) for a · b = 0, b must = 0 since a can't = 0. for b = 0, x must = 0 since a can't = 0.
Remember that a · b = 0 means a or b is the zero vector or a and b are perpendicular to each other. Dot a into both sides of the equation to prove (a).

For (b) use your equation (1) above and the result of (a)
 
  • #3
diazona
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(a) for a · b = 0, b must = 0 since a can't = 0. for b = 0, x must = 0 since a can't = 0.
But remember that you don't know that a · b = 0. You have to prove it.

If the vector identities (the ones you listed under "relevant equations") are what you have to work with, I'd recommend trying to get the original equation (a × x = b) into some form that appears in one of the identities. For instance, try multiplying both sides of a × x = b by something.
 

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