Math
posted by Anonymous .
Given two noncollinear vectors a and b, show that a, axb, and (axb)xa are mutually perpendicular.
Also
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.

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 
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