# Thread: normal line point on surface

1. ## 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!

2. ## Re: normal line point on surface

the normal line to the surface is expressed as:

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

from the given line we have the symmetric equations:

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

which gives us:

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

3. ## 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

4. ## Re: normal line point on surface

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:

$\displaystyle -2$$\displaystyle \spacei\displaystyle + 5$$\displaystyle \space$j$\displaystyle + 3$$\displaystyle \spacek the normal line to the surface coincides with the vector: \displaystyle y$$\displaystyle \space$i$\displaystyle + x$$\displaystyle \spacej\displaystyle - k since the two vectors are parallel, that would make the two lines parallel. So then, \displaystyle y$$\displaystyle \space$i$\displaystyle + x$$\displaystyle \spacej\displaystyle - k\displaystyle = \displaystyle\frac{2}{3}$$\displaystyle \space$i$\displaystyle - \displaystyle\frac{5}{3}$$\displaystyle \spacej\displaystyle -$$\displaystyle \space$k