ohhhhh...but we dnt knw if it passes through j or not so maybe it has a j value ![Huh? ???](https://studentforums.biz/Smileys/default/huh.gif)
consider the whole set up on the
![xy - plane](https://studentforums.biz/cgi-bin/mimetex.cgi? xy - plane )
consider an arbitrary vector and the unit vector
![\vec{\imath}](https://studentforums.biz/cgi-bin/mimetex.cgi?\vec{\imath})
We assume both vectors are parallel.
Lets assume the arbitrary vector is
![a \vec{\imath}](https://studentforums.biz/cgi-bin/mimetex.cgi?a \vec{\imath})
+
![b \vec{\jmath}](https://studentforums.biz/cgi-bin/mimetex.cgi?b \vec{\jmath} )
Gradient of
![\vec{\imath}](https://studentforums.biz/cgi-bin/mimetex.cgi?\vec{\imath})
= Gradient of the other vector
![\frac{\Delta y1}{\Delta x1} = \frac{\Delta y2}{\Delta x2}<br />\frac{0}{1} = \frac{b}{a}<br />\Rightarrow b = 0.<br />](https://studentforums.biz/cgi-bin/mimetex.cgi? \frac{\Delta y1}{\Delta x1} = \frac{\Delta y2}{\Delta x2}<br />\frac{0}{1} = \frac{b}{a}<br />\Rightarrow b = 0.<br />)
Thus if an arbitrary
vector and
![\vec{\imath}](https://studentforums.biz/cgi-bin/mimetex.cgi?\vec{\imath})
are parallel, the co-effecient of j for the arbitrary vector has to equal 0.