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

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SUMMARY

The discussion focuses on calculating the dot product of vectors G and A, where A is defined as A = (a3/l - a1/h) and G is given by G = 2πh(a2 * a3) / (a1 • (a2 * a3)). The conclusion reached is that the dot product results in zero. Participants clarify that the dot product of a vector with itself is not universally 1, emphasizing the need to consider the magnitudes and angles involved in the calculation.

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8614smith
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Homework Statement



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

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 \frac{{{a_3}}}{l} 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?
 
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sorry that first bit should say:

G = 2{\pi}h{\frac{{a_2}x{a_3}}{{a_1}{\bullet}{a_2}x{a_3}}
 
I replaced your formula for G by what you had in your 2nd post.
8614smith said:

Homework Statement



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


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 \frac{{{a_3}}}{l} 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 \cdot v = |u| |v| cos(\theta[\itex]).<br /> <br /> Speaking of vectors, which are the vectors in your problem?
 

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