Dot Product of G and A: Where Do the Denominators Go?

[8614smith]

Homework Statement

what is the dot product of [tex]G{\bullet}A[/tex] where A = [tex]\left(\frac{a_3}{l}-\frac{{a_1}}{h}\right)[/tex] and G = [tex]2{\pi}h{\frac{{a_2}}x{{a_3}}}{{a_1}{\bullet}{{a_2}}x{{a_3}}}[/tex]

Homework Equations

The Attempt at a Solution

The answer is zero and I've got the worked solution infront of me, i just done see where the [tex]\frac{{{a_3}}}{l}[/tex] goes, the dot product of a vector with itself is 1 isn't it? but then where does the denominator go? and what about the denominator of G?

 

[8614smith]

sorry that first bit should say:

G = [tex]2{\pi}h{\frac{{a_2}x{a_3}}{{a_1}{\bullet}{a_2}x{a_3}}[/tex]

 

[Mark44]

I replaced your formula for G by what you had in your 2nd post.

Homework Statement

what is the dot product of [tex]G{\bullet}A[/tex] where A = [tex]\left(\frac{a_3}{l}-\frac{{a_1}}{h}\right)[/tex] and G = [tex]2{\pi}h{\frac{{a_2}x{a_3}}{{a_1}{\bullet}{a_2}x{a_ 3}}[/tex]

Homework Equations

The Attempt at a Solution

The answer is zero and I've got the worked solution infront of me, i just done see where the [tex]\frac{{{a_3}}}{l}[/tex] goes, the dot product of a vector with itself is 1 isn't it? but then where does the denominator go? and what about the denominator of G?

No, the dot product of a vector is not 1 in general. u [itex]\cdot[/itex] v = |u| |v| cos([itex]\theta[\itex]).

Speaking of vectors, which are the vectors in your problem?