Posted by Anonymous on Sunday, March 30, 2008 at 10:57am.
Given two non-collinear vectors a and b, show that a, axb, and (axb)xa are mutually perpendicular.
Express the unit vectors i, j, and k as ordered triplets and show that
i x j = k
what are ordered triplets?
Using components show that
u x v = -v x u for any vectors u and v.
- Math - drwls, Sunday, March 30, 2008 at 11:26am
A cross product is always perpendicular to BOTH vectors, unless it the cross product is zero, which happens only for collinear vectors.
Therefore (a x b) is perpendicular to a.
My statement also proves that (axb)x a is perpendicular to both a and a x b
The "proof" they want you to do involves actually computing the cross product of a x b where a = i and b = j. You will end up proving that the cross product is k. Use these "ordered triplet" designations:
i = (1 0 0)
j = (0 1 0)
k = (0 0 1)
i x j = value of the determinant
|i j k|
|1 0 0|
|0 1 0| = 0i + 0j + k
- Math - drwls, Sunday, March 30, 2008 at 11:35am
Your last question <<Using components show that u x v = -v x u for any vectors u and v.>>
can be done similarly using arbitrary ordered triplet designations for u and v,
u = ai + bj + ck
v = di + ej + fk
where a, b, c, d, e and f are arbitrary constants and then using the determinant formula to compute the cross products. u x v and v x u
u x v =
|i j k|
|a b c|
|d e f|
= (bf - ce)i + (cd - af)j + (ae - bd)k
Now compare that to v x u =
|i j k|
|d e f|
|a b c| = (ec - bf)i, etc
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