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

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
2 replies · 2K views
8614smith
Messages
49
Reaction score
0

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?
 
Physics news on Phys.org
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]
 
I replaced your formula for G by what you had in your 2nd post.
8614smith said:

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]).<br /> <br /> Speaking of vectors, which are the vectors in your problem?[/itex]