Question on dot product of vectors.

[yungman]

[tex] \vec r = \hat x x + \hat y y + \hat z z \;\Rightarrow \;r = \sqrt { x^2+y^2+z^2} \;,\, \hat r= \frac { \hat x x + \hat y y + \hat z z}{ r} [/tex]

I want to find the dot product [tex] (\hat x -\hat r \frac x r) \cdot (\hat x -\hat r \frac x r) [/tex]

1) [tex] \hat x -\hat r \frac x r = \hat x - \frac { (\hat x x + \hat y y + \hat z z) x }{ r^2} = \frac { \hat x (y^2+z^2) - \hat y xy - \hat z xz}{r^2} \;\Rightarrow\; (\hat x -\hat r \frac x r) \cdot (\hat x -\hat r \frac x r) = \frac {(y^2+z^2)^2+x^2y^2+x^2z^2}{r^4}= \frac {(y^2+z^2)(x^2+y^2+z^2)}{r^4}=\frac {(y^2+z^2)}{r^2}[/tex]




2) But if I just blind do the dot product:

[tex] (\hat x -\hat r \frac x r) \cdot (\hat x -\hat r \frac x r) = 1 + \frac {x^2}{r^2} [/tex]

Here, all I did is [itex] \hat x \cdot \hat x \;\hbox { and } \hat r \cdot \hat r [/itex].




Is it true for dot product, only independent variable can dot together, [itex] \hat r[/itex] is a dependent variable of [itex] \hat x[/itex] so I cannot use the 2) method to perform dot product. Is this true?

Thanks

Alan

 

[piyushkumar]

when u open the bracket directly (as done in point 2), you get three terms, one of x_cap.x_cap, one of r_cap.r_cap, which you have written write, but term of x_cap.r_cap you have missed.

since r_cap is (x x_cap + y y_cap + z z_cap)/[itex]\sqrt{}(x*x+y*y+z*z)[/itex], so r_cap.x_cap is not zero, as your second method is actually doing

 

[yungman]

when u open the bracket directly (as done in point 2), you get three terms, one of x_cap.x_cap, one of r_cap.r_cap, which you have written write, but term of x_cap.r_cap you have missed.

since r_cap is (x x_cap + y y_cap + z z_cap)/[itex]\sqrt{}(x*x+y*y+z*z)[/itex], so r_cap.x_cap is not zero, as your second method is actually doing

I got it. Thanks.

Alan