How do I prove that A^B is orthogonal to A?

[EmmaLemming]

If A = (2,-2,1) and B = (2, 0, -1)

show by explicit calculation that;

i) A^B is orthogonal to A

ii) (A^B)^B lies in the same plane as A and B by expressing it as a linear combination of A and B

I'm using;
A^B = |A||B|sin θ

I know that when you do the cross product of two vectors the result will be a vector that is perpendicular to both and I can draw a diagram to demonstrate. However I can't show it by explicit calculation.

I thought maybe if I could prove that the angle between them was ∏/2...

 

[phyzguy]

When you write A^B = |A||B|sin θ, that only gives the magnitude of A^B. You need to include the direction information as well. In three dimensions the dual of A^B is a vector, which is given by the cross product. So, for part A, try forming the dot product of (AxB) with A and see what this gives.

 

[Mark44]

If A = (2,-2,1) and B = (2, 0, -1)

show by explicit calculation that;

i) A^B is orthogonal to A

ii) (A^B)^B lies in the same plane as A and B by expressing it as a linear combination of A and B

I'm using;
A^B = |A||B|sin θ

I know that when you do the cross product of two vectors the result will be a vector that is perpendicular to both and I can draw a diagram to demonstrate. However I can't show it by explicit calculation.

I thought maybe if I could prove that the angle between them was ∏/2...

Since you're talking about the cross product, why don't you write it as A x B?