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.

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.

- 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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