# normal line point on surface

• Apr 28th 2013, 05:06 PM
topsaa
normal line point on surface
Would anyone know how to begin a problem where a function z(x,y) is given and x(t),y(t),z(t).

Thanks!
• Apr 28th 2013, 07:30 PM
jpritch422
Re: normal line point on surface
the normal line to the surface is expressed as:

N $= \displaystyle\frac{\partial{z}}{\partial{x}}$ $\space$i $+ \displaystyle\frac{\partial{z}}{\partial{y}}$ $\space$j $-$ $\space$k

$\displaystyle\frac{\partial{z}}{\partial{x}} = y$; $\space$ $\displaystyle\frac{\partial{z}}{\partial{y}} = x$

N $= y$ $\space$i $+ x$ $\space$j $-$ $\space$k

from the given line we have the symmetric equations:

$\displaystyle\frac{x - 3}{-2} = \displaystyle\frac{y - 4}{5} = \displaystyle\frac{z - 3}{3}$

which gives us:

$\space$ $-2$i $\space$ $+$ $\space$ $5$j $\space$ $+$ $\space$ $3$k
• Apr 28th 2013, 08:14 PM
topsaa
Re: normal line point on surface
so we would solve for where <y,x,-1> is equal to <-2,5,3>, wouldn't that be never because of k, -1 cannot equal 3
• Apr 28th 2013, 08:55 PM
jpritch422
Re: normal line point on surface
Quote:

Originally Posted by topsaa
so we would solve for where <y,x,-1> is equal to <-2,5,3>, wouldn't that be never because of k, -1 cannot equal 3

what this is saying is that the given line coincides with vector:

$-2$ $\space$i $+ 5$ $\space$j $+ 3$ $\space$k

the normal line to the surface coincides with the vector:

$y$ $\space$i $+ x$ $\space$j $-$k

since the two vectors are parallel, that would make the two lines parallel. So then,

$y$ $\space$i $+ x$ $\space$j $-$k $= \displaystyle\frac{2}{3}$ $\space$i $- \displaystyle\frac{5}{3}$ $\space$j $-$ $\space$k